PAPER: TEXAS INSTRUMENTS PLACEMENT PAPER - 08 SEP 2005 ( TECHNICAL- C, C++)

 

1. Can we declare a static function as virtual?

Ans: No. The virtual function mechanism is used on the specific object that determines which virtual function to call. Since the static functions are not any way related to objects, they cannot be declared as virtual.

 

2. Can user-defined object be declared as static data member of another class?

Ans: Yes. The following code shows how to initialize a user-defined object.

#include

class test

{

int i ;

public :

test ( int ii = 0 )

{

i = ii ;

}

} ;

class sample

{

static test s ;

} ;

test sample::s ( 26 ) ;

Here we have initialized the object s by calling the one-argument constructor. We

can use the same convention to initialize the object by calling multiple-argument constructor.

 

3. What is forward referencing and when should it be used?

Ans: Consider the following program:

class test

{

public :

friend void fun ( sample, test ) ;

} ;

class sample

{

public :

friend void fun ( sample, test ) ;

} ;

void fun ( sample s, test t )

{

// code

}

void main( )

{

sample s ;

test t ;

fun ( s, t ) ;

}

This program would not compile. It gives an error that sample is undeclared identifier in the statement friend void fun ( sample, test ) ; of the class test. This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample. Forward referencing is generally required when we make a class or a function as a friend.

 

4. The istream_withassign class has been derived from the istream class and overloaded assignment operator has been added to it. The _withassign classes are much like their base classes except that they include overloaded assignment operators. Using these operators the objects of the _withassign classes can be copied. The istream, ostream, and iostream classes are made uncopyable by making their overloaded copy constructor and assignment operators private.

 

5. How do I write my own zero-argument manipulator that should work same as hex?

Ans: This is shown in following program.

#include

ostream& myhex ( ostream &o )

{

o.setf ( ios::hex) ;

return o ;

}

void main( )

{

cout << endl << myhex << 2000 ;

}

 

6.We all know that a const variable needs to be initialized at the time of declaration. Then how come the program given below runs properly even when we have not initialized p?

#include

void main( )

{

const char *p ;

p = "A const pointer" ;

cout << p ;

}

Ans: The output of the above program is 'A const pointer'. This is because in this program p is declared as 'const char*' which means that value stored at p will be constant and not p and so the program works properly

 

7. How do I refer to a name of class or function that is defined within a namespace?

Ans: There are two ways in which we can refer to a name of class or function that is defined within a namespace: Using scope resolution operator through the using keyword. This is shown in following example:

 

namespace name1

{

class sample1

{

// code

} ;

}

namespace name2

{

class sample2

{

// code

} ;

}

using namespace name2 ;

void main( )

{

name1::sample1 s1 ;

sample2 s2 ;

}

Here, class sample1 is referred using the scope resolution operator. On the other hand we can directly refer to class sample2 because of the statement using namespace name2 ; the using keyword declares all the names in the namespace to be in the current scope. So we can use the names without any qualifiers.

 

8. While overloading a binary operator can we provide default values?

Ans: No!. This is because even if we provide the default arguments to the parameters of the overloaded operator function we would end up using the binary operator incorrectly. This is explained in the following example:

 

sample operator + ( sample a, sample b = sample (2, 3.5f ) )

{

}

void main( )

{

sample s1, s2, s3 ;

s3 = s1 + ; // error

}

 

9. How do I carry out conversion of one object of user-defined type to another?

Ans: To perform conversion from one user-defined type to another we need to provide conversion function. Following program demonstrates how to provide such conversion function.

class circle

{

private :

int radius ;

public:

circle ( int r = 0 )

{

radius = r ;

}

} ;

class rectangle

{

private :

int length, breadth ;

public :

rectangle( int l, int b )

{

length = l ;

breadth = b ;

}

operator circle( )

{

return circle ( length ) ;

}

} ;

void main( )

{

rectangle r ( 20, 10 ) ;

circle c;

c = r ;

}

Here, when the statement c = r ; is executed the compiler searches for an overloaded assignment operator in the class circle which accepts the object of type rectangle. Since there is no such overloaded assignment operator, the conversion operator function that converts the rectangle object to the circle object is searched in the rectangle class. We have provided such a conversion function in the rectangle class. This conversion operator function returns a circle object. By default conversion operators have the name and return type same as the object type to which it converts to. Here the type of the object is circle and hence the name of the operator function as well as the return type is circle.

 

10. How do I write code that allows to create only one instance of a class?

Ans: This is shown in following code snippet.

 

#include

class sample

{

static sample *ptr ;

private:

sample( )

{

}

public:

static sample* create( )

{

if ( ptr == NULL )

ptr = new sample ;

return ptr ;

}

} ;

sample *sample::ptr = NULL ;

void main( )

{

sample *a = sample::create( ) ;

sample *b = sample::create( ) ;

}

Here, the class sample contains a static data member ptr, which is a pointer

to the object of same class. The constructor is private which avoids us from creating objects outside the class. A static member function called create( ) is used to create an object of the class. In this function the condition is checked whether or not ptr is NULL, if it is then an object is created dynamically and its address collected in ptr is returned. If ptr is not NULL, then the same address is returned. Thus, in main( ) on execution of the first statement one object of sample gets created whereas on execution of second statement, b holds the address of the first object. Thus, whatever number of times you call create( ) function, only one object of sample class will be available.

 

11. How do I write code to add functions, which would work as get and put properties of a class?

Ans: This is shown in following code.

#include

class sample

{

int data ;

public:

__declspec ( property ( put = fun1, get = fun2 ) )

int x ;

void fun1 ( int i )

{

if ( i < 0 )

data = 0 ;

else

data = i ;

}

int fun2( )

{

return data ;

}

} ;

void main( )

{

sample a ;

a.x = -99 ;

cout << a.x ;

}

Here, the function fun1( ) of class sample is used to set the given integer value into data, whereas fun2( ) returns the current value of data. To set these functions as properties of a class we have given the statement as shown below:

__declspec ( property ( put = fun1, get = fun2 )) int x ;

 

As a result, the statement a.x = -99 ; would cause fun1( ) to get called to set the value in data. On the other hand, the last statement would cause fun2( ) to get called to return the value of data.

 

12. How do I write code to make an object work like a 2-D array?

Ans: Take a look at the following program.

#include

class emp

{

public :

int a[3][3] ;

emp( )

{

int c = 1 ;

for ( int i = 0 ; i <= 2 ; i++ )

{

for ( int j = 0 ; j <= 2 ; j++ )

{

a[i][j] = c ;

c++ ;

}

}

}

int* operator[] ( int i )

{

return a[i] ;

}

} ;

void main( )

{

emp e ;

cout << e[0][1] ;

}

The class emp has an overloaded operator [ ] function. It takes one argument an integer representing an array index and returns an int pointer. The statement cout << e[0][1] ; would get converted into a call to the overloaded [ ] function as e.operator[ ] ( 0 ). 0 would get collected in i. The function would return a[i] that represents the base address of the zeroeth row. Next the statement would get expanded as base address of zeroeth row[1] that can be further expanded as *( base address + 1 ). This gives us a value in zeroth row and first column.

 

13. What are formatting flags in ios class?

Ans: The ios class contains formatting flags that help users to format the stream data. Formatting flags are a set of enum definitions. There are two types of formatting flags:

On/Off flags :Flags that work in-group The On/Off flags are turned on using the setf( ) function and are turned off using the unsetf( ) function. To set the On/Off flags, the one argument setf( ) function is used. The flags working in groups are set through the two-argument setf( ) function. For example, to left justify a string we can set the flag as,

cout.setf ( ios::left ) ;

cout << "KICIT Nagpur" ;

To remove the left justification for subsequent output we can say,

cout.unsetf ( ios::left ) ;

The flags that can be set/unset include skipws, showbase, showpoint,

uppercase, showpos, unitbuf and stdio. The flags that work in a group can have only one of these flags set at a time.

 

14. What is the purpose of ios::basefield in the following statement?

cout.setf ( ios::hex, ios::basefield ) ;

Ans: This is an example of formatting flags that work in a group. There is a flag for each numbering system (base) like decimal, octal and hexadecimal. Collectively, these flags are referred to as basefield and are specified by ios::basefield flag. We can have only one of these flags on at a time. If we set the hex flag as setf ( ios::hex ) then we will set the hex bit but we won't clear the dec bit resulting in undefined behavior. The solution is to call setf( ) as setf ( ios::hex, ios::basefield ). This call first clears all the bits and then sets the hex bit.

 

15. Can we get the value of ios format flags?

Ans: Yes! The ios::flags( ) member function gives the value format flags. This function takes no arguments and returns a long ( typedefed to fmtflags) that contains the current format flags.

 

16. Is there any function that can skip certain number of characters present in the input stream?

Ans: Yes! This can be done using cin::ignore( ) function. The prototype of this function is as shown below:

istream& ignore ( int n = 1, int d =EOF ) ;

Sometimes it happens that some extra characters are left in the input stream while taking the input such as, the ?\n? (Enter) character. This extra character is then passed to the next input and may pose problem.

To get rid of such extra characters the cin::ignore( ) function is used. This is equivalent to fflush ( stdin ) used in C language. This function ignores the first n characters (if present) in the input stream, stops if delimiter d is encountered.

 

17. Write a program that implements a date class containing day, month and year as data members. Implement assignment operator and copy constructor in this class.

Ans: This is shown in following program:

#include

class date

{

private :

int day ;

int month ;

int year ;

public :

date ( int d = 0, int m = 0, int y = 0 )

{

day = d ;

month = m ;

year = y ;

}

// copy constructor

date ( date &d )

{

day = d.day ;

month = d.month ;

year = d.year ;

}

// an overloaded assignment operator

date operator = ( date d )

{

day = d.day ;

month = d.month ;

year = d.year ;

return d ;

}

void display( )

{

cout << day << "/" << month << "/" << year ;

}

} ;

void main( )

{

date d1 ( 25, 9, 1979 ) ;

date d2 = d1 ;

date d3 ;

d3 = d2 ;

d3.display( ) ;

}

 

18. When should I use unitbuf flag?

Ans: The unit buffering (unitbuf) flag should be turned on when we want to ensure that each character is output as soon as it is inserted into an output stream. The same can be done using unbuffered output but unit buffering provides a better performance than the unbuffered output.

 

19.What are manipulators?

Ans: Manipulators are the instructions to the output stream to modify the output in various ways. The manipulators provide a clean and easy way for formatted output in comparison to the formatting flags of the ios class. When manipulators are used, the formatting instructions are inserted directly into the stream. Manipulators are of two types, those that take an argument and those that don?t.

 

20. What is the difference between the manipulator and setf( ) function?

Ans: The difference between the manipulator and setf( ) function are as follows:

 

The setf( ) function is used to set the flags of the ios but manipulators directly insert the formatting instructions into the stream. We can create user-defined manipulators but setf( ) function uses data members of ios class only. The flags put on through the setf( ) function can be put off through unsetf( ) function. Such flexibility is not available with manipulators.

 

21. How do I get the current position of the file pointer?

Ans: We can get the current position of the file pointer by using the tellp( ) member function of ostream class or tellg( ) member function of istream class. These functions return (in bytes) positions of put pointer and get pointer respectively.

 

22. What are put and get pointers?

Ans: These are the long integers associated with the streams. The value present in the put pointer specifies the byte number in the file from where next write would take place in the file. The get pointer specifies the byte number in the file from where the next reading should take place.

 

23. What do the nocreate and noreplace flag ensure when they are used for opening a file?

Ans: nocreate and noreplace are file-opening modes. A bit in the ios class defines these modes. The flag nocreate ensures that the file must exist before opening it. On the other hand the flag noreplace ensures that while opening a file for output it does not get overwritten with new one unless ate or app is set. When the app flag is set then whatever we write gets appended to the existing file. When ate flag is set we can start reading or writing at the end of existing file.

 

24. What is the limitation of cin while taking input for character array?

Ans: To understand this consider following statements,

char str[5] ;

cin >> str ;

While entering the value for str if we enter more than 5 characters then there is no provision in cin to check the array bounds. If the array overflows, it may be dangerous. This can be avoided by using get( ) function. For example, consider following statement,

cin.get ( str, 5 ) ;

On executing this statement if we enter more than 5 characters, then get( ) takes only first five characters and ignores rest of the characters. Some more variations of get( ) are available, such as shown below:

get ( ch ) ? Extracts one character only

get ( str, n ) ? Extracts up to n characters into str

get ( str, DELIM ) ? Extracts characters into array str until specified delimiter (such as '\n'). Leaves delimiting character in stream.

get ( str, n, DELIM ) ? Extracts characters into array str until n characters or DELIM character, leaving delimiting character in stream.

 

25. What is the purpose of istream class?

Ans: The istream class performs activities specific to input. It is derived from the ios class. The most commonly used member function of this class is the overloaded >> operator which can extract values of all basic types. We can extract even a string using this operator.

 

26. Would the following code work?

#include

void main( )

{

ostream o ;

o << "Dream. Then make it happen!" ;

}

Ans: No! This is because we cannot create an object of the ostream class since its constructor and copy constructor are declared private.

 

27. Can we use this pointer inside static member function?

Ans: No! The this pointer cannot be used inside a static member function. This is because a static member function is never called through an object.

 

28. What is strstream?

Ans: strstream is a type of input/output stream that works with the memory. It allows using section of the memory as a stream object. These streams provide the classes that can be used for storing the stream of bytes into memory. For example, we can store integers, floats and strings as a stream of bytes. There are several classes that implement this in-memory formatting. The class ostrstream derived from ostream is used when output is to be sent to memory, the class istrstream derived from istream is used when input is taken from memory and strstream class derived from iostream is used for memory objects that do both input and output. Ans: When we want to retrieve the streams of bytes from memory we can use istrestream. The following example shows the use of istrstream class.

#include

void main( )

{

int age ;

float salary ;

char name[50] ;

char str[] = "22 12004.50 K. Vishwanatth" ;

istrstream s ( str ) ;

s >> age >> salary >> name ;

cout << age << endl << salary << endl << name ;

cout << endl << s.rdbuf( ) ;

}

Here, s is the object of the class istrstream. When we are creating the object s, the constructor of istrstream gets called that receives a pointer to the zero terminated character array str. The statement s >> age >> salary >> name ; extracts the age, salary and the name from the istrstream object s. However, while extracting the name, only the first word of name gets extracted. The balance is extracted using rdbuf( ).

 

29. When the constructor of a base class calls a virtual function, why doesn't the override function of the derived class gets called?

Ans: While building an object of a derived class first the constructor of the base class and then the constructor of the derived class gets called. The object is said an immature object at the stage when the constructor of base class is called. This object will be called a matured object after the execution of the constructor of the derived class. Thus, if we call a virtual function when an object is still immature, obviously, the virtual function of the base class would get called. This is illustrated in the following example.

#include

class base

{

protected :

int i ;

public :

base ( int ii = 0 )

{

i = ii ;

show( ) ;

}

virtual void show( )

{

cout << "base's show( )" << endl ;

}

} ;

class derived : public base

{

private :

int j ;

public :

derived ( int ii, int jj = 0 ) : base ( ii )

{

j = jj ;

show( ) ;

}

void show( )

{

cout << "derived's show( )" << endl ;

}

} ;

 

void main( )

{

derived dobj ( 20, 5 ) ;

}

The output of this program would be:

base's show( )

derived's show( )

 

30. Can I have a reference as a data member of a class? If yes, then how do I initialise it?

Ans: Yes, we can have a reference as a data member of a class. A reference as a data member of a class is initialized in the initialization list of the constructor. This is shown in following program.

#include

class sample

{

private :

int& i ;

public :

sample ( int& ii ) : i ( ii )

{

}

void show( )

{

cout << i << endl ;

}

} ;

void main( )

{

int j = 10 ;

sample s ( j ) ;

s.show( ) ;

}

Here, i refers to a variable j allocated on the stack. A point to note here is that we cannot bind a reference to an object passed to the constructor as a value. If we do so, then the reference i would refer to the function parameter (i.e. parameter ii in the constructor), which would disappear as soon as the function returns, thereby creating a situation of dangling reference.

 

31. Why does the following code fail?

#include

class sample

{

private :

char *str ;

public :

sample ( char *s )

{

strcpy ( str, s ) ;

}

~sample( )

{

delete str ;

}

} ;

void main( )

{

sample s1 ( "abc" ) ;

}

Ans: Here, through the destructor we are trying to deal locate memory, which has been allocated statically. To remove an exception, add following statement to the constructor.

sample ( char *s )

{

str = new char[strlen(s) + 1] ;

strcpy ( str, s ) ;

}

Here, first we have allocated memory of required size, which then would get deal located through the destructor.

 

32. assert( ) macro...

We can use a macro called assert( ) to test for conditions that should not occur in a code. This macro expands to an if statement. If test evaluates to 0, assert prints an error message and calls abort to abort the program.

#include

#include

void main( )

{

int i ;

cout << "\nEnter an integer: " ;

cin >> i ;

assert ( i >= 0 ) ;

cout << i << endl ;

}

 

33. Why it is unsafe to deal locate the memory using free( ) if it has been allocated using new?

Ans: This can be explained with the following example:

#include

class sample

{

int *p ;

public :

sample( )

{

p = new int ;

}

~sample( )

{

delete p ;

}

} ;

void main( )

{

sample *s1 = new sample ;

free ( s1 ) ;

sample *s2 = ( sample * ) malloc ( sizeof ( sample ) ) ;

delete s2 ;

}

The new operator allocates memory and calls the constructor. In the constructor we have allocated memory on heap, which is pointed to by p. If we release the object using the free( ) function the object would die but the memory allocated in the constructor would leak. This is because free( ) being a C library function does not call the destructor where we have deal located the memory.

 

As against this, if we allocate memory by calling malloc( ) the constructor would not get called. Hence p holds a garbage address. Now if the memory is deal located using delete, the destructor would get called where we have tried to release the memory pointed to by p. Since p contains garbage this may result in a runtime error.

 

34. Can we distribute function templates and class templates in object libraries?

Ans: No! We can compile a function template or a class template into object code (.obj file). The code that contains a call to the function template or the code that creates an object from a class template can get compiled. This is because the compiler merely checks whether the call matches the declaration (in case of function template) and whether the object definition matches class declaration (in case of class template). Since the function template and the class template definitions are not found, the compiler leaves it to the linker to restore this. However, during linking, linker doesn't find the matching definitions for the function call or a matching definition for object creation. In short the expanded versions of templates are not found in the object library. Hence the linker reports error.

 

35. What is the difference between an inspector and a mutator ?

Ans: An inspector is a member function that returns information about an object's state (information stored in object's data members) without changing the object's state. A mutator is a member function that changes the state of an object. In the class Stack given below we have defined a mutator and an inspector.

class Stack

{

public :

int pop( ) ;

int getcount( ) ;

}

In the above example, the function pop( ) removes top element of stack thereby changing the state of an object. So, the function pop( ) is a mutator. The function getcount( ) is an inspector because it simply counts the number of elements in the stack without changing the stack.

 

36. Namespaces:

The C++ language provides a single global namespace. This can cause problems with global name clashes. For instance, consider these two C++ header files:

// file1.h

float f ( float, int ) ;

class sample { ... } ;

// file2.h

class sample { ... } ;

With these definitions, it is impossible to use both header files in a single program; the sample classes will clash. A namespace is a declarative region that attaches an additional identifier to any names declared inside it. The additional identifier thus avoids the possibility that a name will conflict with names declared elsewhere in the program. It is possible to use the same name in separate namespaces without conflict even if the names appear in the same translation unit. As long as they appear in separate namespaces, each name will be unique because of the addition of the namespace identifier. For example:

// file1.h

namespace file1

{

float f ( float, int ) ;

class sample { ... } ;

}

// file2.h

namespace file2

{

class sample { ... } ;

}

 

Now the class names will not clash because they become file1::sample and file2::sample, respectively.

 

37. What would be the output of the following program?

#include

class user

{

int i ;

float f ;

char c ;

public :

void displaydata( )

{

cout << endl << i << endl << f << endl << c ;

}

} ;

void main( )

{

cout << sizeof ( user ) ;

user u1 ;

cout << endl << sizeof ( u1 ) ;

u1.displaydata( ) ;

}

Ans: The output of this program would be,

9 or 7

9 or 7

Garbage

Garbage

Garbage

Since the user class contains three elements, int, float and char its size would be 9 bytes (int-4, float-4, char-1) under Windows and 7 bytes (int-2, float-4, char-1) under DOS. Second output is again the same because u1 is an object of the class user. Finally three garbage values are printed out because i, f and c are not initialized anywhere in the program.

 

Note that if you run this program you may not get the answer shown here. This is because packing is done for an object in memory to increase the access efficiency. For example, under DOS, the object would be aligned on a 2-byte boundary. As a result, the size of the object would be reported as 6 bytes. Unlike this, Windows being a 32-bit OS the object would be aligned on a 4-byte boundary. Hence the size of the object would be reported as 12 bytes. To force the alignment on a 1-byte boundary, write the following statement before the class declaration.

#pragma pack ( 1 )

 

38. Write a program that will convert an integer pointer to an integer and vice-versa.

Ans: The following program demonstrates this.

#include

void main( )

{

int i = 65000 ;

int *iptr = reinterpret_cast ( i ) ;

cout << endl << iptr ;

iptr++ ;

cout << endl << iptr ;

i = reinterpret_cast ( iptr ) ;

cout << endl << i ;

i++ ;

cout << endl << i ;

}

 

39. What is a const_cast?

Ans. The const_cast is used to convert a const to a non-const. This is shown in the following

program:

#include

void main( )

{

const int a = 0 ;

int *ptr = ( int * ) &a ; //one way

ptr = const_cast_ ( &a ) ; //better way

}

Here, the address of the const variable a is assigned to the pointer to a non-const variable. The const_cast is also used when we want to change the data members of a class inside the const member functions. The following code snippet shows this:

class sample

{

private:

int data;

public:

void func( ) const

{

(const_cast (this))->data = 70 ;

}

} ;

 

40. What is forward referencing and when should it be used?

Ans: Forward referencing is generally required when we make a class or a function as a friend.

Consider following program:

class test

{

public:

friend void fun ( sample, test ) ;

} ;

class sample

{

public:

friend void fun ( sample, test ) ;

} ;

void fun ( sample s, test t )

{

// code

}

void main( )

{

sample s ;

test t ;

fun ( s, t ) ;

}

On compiling this program it gives error on the following statement of test class. It gives an error that sample is undeclared identifier. friend void fun ( sample, test );

This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample.

class sample ;

 

41. How would you give an alternate name to a namespace?

Ans: An alternate name given to namespace is called a namespace-alias. namespace-alias is generally used to save the typing effort when the names of namespaces are very long or complex. The following syntax is used to give an alias to a namespace.

namespace myname = my_old_very_long_name ;

 

42. Using a smart pointer can we iterate through a container?

Ans: Yes. A container is a collection of elements or objects. It helps to properly organize and store the data. Stacks, linked lists, arrays are examples of containers. Following program shows how to iterate through a container using a smart pointer.

#include

class smartpointer

{

private :

int *p ; // ordinary pointer

public :

smartpointer ( int n )

{

p = new int [ n ] ;

int *t = p ;

for ( int i = 0 ; i <= 9 ; i++ )

*t++ = i * i ;

}

int* operator ++ ( int )

{

return p++ ;

}

int operator * ( )

{

return *p ;

}

} ;

void main( )

{

smartpointer sp ( 10 ) ;

for ( int i = 0 ; i <= 9 ; i++ )

cout << *sp++ << endl ;

}

Here, sp is a smart pointer. When we say *sp, the operator * ( ) function gets called. It returns the integer being pointed to by p. When we say sp++ the operator ++ ( ) function gets called. It increments p to point to The next element in the array and then returns the address of this new location.

 

43. Can objects read and write themselves?

Ans: Yes! This can be explained with the help of following example:

#include

#include

class employee

{

private :

char name [ 20 ] ;

int age ;

float salary ;

public :

void getdata( )

{

cout << "Enter name, age and salary of employee : " ;

cin >> name >> age >> salary ;

}

void store( )

{

ofstream file ;

file.open ( "EMPLOYEE.DAT", ios::app | ios::binary ) ;

file.write ( ( char * ) this, sizeof ( *this ) ) ;

file.close( ) ;

}

void retrieve ( int n )

{

ifstream file ;

file.open ( "EMPLOYEE.DAT", ios:

 

PAPER: TEXAS INSTRUMENTS PLACEMENT PAPER - 08 SEP 2005 ( TECHNICAL- C, C++)

 

1. Can we declare a static function as virtual?

Ans: No. The virtual function mechanism is used on the specific object that determines which virtual function to call. Since the static functions are not any way related to objects, they cannot be declared as virtual.

 

2. Can user-defined object be declared as static data member of another class?

Ans: Yes. The following code shows how to initialize a user-defined object.

#include

class test

{

int i ;

public :

test ( int ii = 0 )

{

i = ii ;

}

} ;

class sample

{

static test s ;

} ;

test sample::s ( 26 ) ;

Here we have initialized the object s by calling the one-argument constructor. We

can use the same convention to initialize the object by calling multiple-argument constructor.

 

3. What is forward referencing and when should it be used?

Ans: Consider the following program:

class test

{

public :

friend void fun ( sample, test ) ;

} ;

class sample

{

public :

friend void fun ( sample, test ) ;

} ;

void fun ( sample s, test t )

{

// code

}

void main( )

{

sample s ;

test t ;

fun ( s, t ) ;

}

This program would not compile. It gives an error that sample is undeclared identifier in the statement friend void fun ( sample, test ) ; of the class test. This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample. Forward referencing is generally required when we make a class or a function as a friend.

 

4. The istream_withassign class has been derived from the istream class and overloaded assignment operator has been added to it. The _withassign classes are much like their base classes except that they include overloaded assignment operators. Using these operators the objects of the _withassign classes can be copied. The istream, ostream, and iostream classes are made uncopyable by making their overloaded copy constructor and assignment operators private.

 

5. How do I write my own zero-argument manipulator that should work same as hex?

Ans: This is shown in following program.

#include

ostream& myhex ( ostream &o )

{

o.setf ( ios::hex) ;

return o ;

}

void main( )

{

cout << endl << myhex << 2000 ;

}

 

6.We all know that a const variable needs to be initialized at the time of declaration. Then how come the program given below runs properly even when we have not initialized p?

#include

void main( )

{

const char *p ;

p = "A const pointer" ;

cout << p ;

}

Ans: The output of the above program is 'A const pointer'. This is because in this program p is declared as 'const char*' which means that value stored at p will be constant and not p and so the program works properly

 

7. How do I refer to a name of class or function that is defined within a namespace?

Ans: There are two ways in which we can refer to a name of class or function that is defined within a namespace: Using scope resolution operator through the using keyword. This is shown in following example:

 

namespace name1

{

class sample1

{

// code

} ;

}

namespace name2

{

class sample2

{

// code

} ;

}

using namespace name2 ;

void main( )

{

name1::sample1 s1 ;

sample2 s2 ;

}

Here, class sample1 is referred using the scope resolution operator. On the other hand we can directly refer to class sample2 because of the statement using namespace name2 ; the using keyword declares all the names in the namespace to be in the current scope. So we can use the names without any qualifiers.

 

8. While overloading a binary operator can we provide default values?

Ans: No!. This is because even if we provide the default arguments to the parameters of the overloaded operator function we would end up using the binary operator incorrectly. This is explained in the following example:

 

sample operator + ( sample a, sample b = sample (2, 3.5f ) )

{

}

void main( )

{

sample s1, s2, s3 ;

s3 = s1 + ; // error

}

 

9. How do I carry out conversion of one object of user-defined type to another?

Ans: To perform conversion from one user-defined type to another we need to provide conversion function. Following program demonstrates how to provide such conversion function.

class circle

{

private :

int radius ;

public:

circle ( int r = 0 )

{

radius = r ;

}

} ;

class rectangle

{

private :

int length, breadth ;

public :

rectangle( int l, int b )

{

length = l ;

breadth = b ;

}

operator circle( )

{

return circle ( length ) ;

}

} ;

void main( )

{

rectangle r ( 20, 10 ) ;

circle c;

c = r ;

}

Here, when the statement c = r ; is executed the compiler searches for an overloaded assignment operator in the class circle which accepts the object of type rectangle. Since there is no such overloaded assignment operator, the conversion operator function that converts the rectangle object to the circle object is searched in the rectangle class. We have provided such a conversion function in the rectangle class. This conversion operator function returns a circle object. By default conversion operators have the name and return type same as the object type to which it converts to. Here the type of the object is circle and hence the name of the operator function as well as the return type is circle.

 

10. How do I write code that allows to create only one instance of a class?

Ans: This is shown in following code snippet.

 

#include

class sample

{

static sample *ptr ;

private:

sample( )

{

}

public:

static sample* create( )

{

if ( ptr == NULL )

ptr = new sample ;

return ptr ;

}

} ;

sample *sample::ptr = NULL ;

void main( )

{

sample *a = sample::create( ) ;

sample *b = sample::create( ) ;

}

Here, the class sample contains a static data member ptr, which is a pointer

to the object of same class. The constructor is private which avoids us from creating objects outside the class. A static member function called create( ) is used to create an object of the class. In this function the condition is checked whether or not ptr is NULL, if it is then an object is created dynamically and its address collected in ptr is returned. If ptr is not NULL, then the same address is returned. Thus, in main( ) on execution of the first statement one object of sample gets created whereas on execution of second statement, b holds the address of the first object. Thus, whatever number of times you call create( ) function, only one object of sample class will be available.

 

11. How do I write code to add functions, which would work as get and put properties of a class?

Ans: This is shown in following code.

#include

class sample

{

int data ;

public:

__declspec ( property ( put = fun1, get = fun2 ) )

int x ;

void fun1 ( int i )

{

if ( i < 0 )

data = 0 ;

else

data = i ;

}

int fun2( )

{

return data ;

}

} ;

void main( )

{

sample a ;

a.x = -99 ;

cout << a.x ;

}

Here, the function fun1( ) of class sample is used to set the given integer value into data, whereas fun2( ) returns the current value of data. To set these functions as properties of a class we have given the statement as shown below:

__declspec ( property ( put = fun1, get = fun2 )) int x ;

 

As a result, the statement a.x = -99 ; would cause fun1( ) to get called to set the value in data. On the other hand, the last statement would cause fun2( ) to get called to return the value of data.

 

12. How do I write code to make an object work like a 2-D array?

Ans: Take a look at the following program.

#include

class emp

{

public :

int a[3][3] ;

emp( )

{

int c = 1 ;

for ( int i = 0 ; i <= 2 ; i++ )

{

for ( int j = 0 ; j <= 2 ; j++ )

{

a[i][j] = c ;

c++ ;

}

}

}

int* operator[] ( int i )

{

return a[i] ;

}

} ;

void main( )

{

emp e ;

cout << e[0][1] ;

}

The class emp has an overloaded operator [ ] function. It takes one argument an integer representing an array index and returns an int pointer. The statement cout << e[0][1] ; would get converted into a call to the overloaded [ ] function as e.operator[ ] ( 0 ). 0 would get collected in i. The function would return a[i] that represents the base address of the zeroeth row. Next the statement would get expanded as base address of zeroeth row[1] that can be further expanded as *( base address + 1 ). This gives us a value in zeroth row and first column.

 

13. What are formatting flags in ios class?

Ans: The ios class contains formatting flags that help users to format the stream data. Formatting flags are a set of enum definitions. There are two types of formatting flags:

On/Off flags :Flags that work in-group The On/Off flags are turned on using the setf( ) function and are turned off using the unsetf( ) function. To set the On/Off flags, the one argument setf( ) function is used. The flags working in groups are set through the two-argument setf( ) function. For example, to left justify a string we can set the flag as,

cout.setf ( ios::left ) ;

cout << "KICIT Nagpur" ;

To remove the left justification for subsequent output we can say,

cout.unsetf ( ios::left ) ;

The flags that can be set/unset include skipws, showbase, showpoint,

uppercase, showpos, unitbuf and stdio. The flags that work in a group can have only one of these flags set at a time.

 

14. What is the purpose of ios::basefield in the following statement?

cout.setf ( ios::hex, ios::basefield ) ;

Ans: This is an example of formatting flags that work in a group. There is a flag for each numbering system (base) like decimal, octal and hexadecimal. Collectively, these flags are referred to as basefield and are specified by ios::basefield flag. We can have only one of these flags on at a time. If we set the hex flag as setf ( ios::hex ) then we will set the hex bit but we won't clear the dec bit resulting in undefined behavior. The solution is to call setf( ) as setf ( ios::hex, ios::basefield ). This call first clears all the bits and then sets the hex bit.

 

15. Can we get the value of ios format flags?

Ans: Yes! The ios::flags( ) member function gives the value format flags. This function takes no arguments and returns a long ( typedefed to fmtflags) that contains the current format flags.

 

16. Is there any function that can skip certain number of characters present in the input stream?

Ans: Yes! This can be done using cin::ignore( ) function. The prototype of this function is as shown below:

istream& ignore ( int n = 1, int d =EOF ) ;

Sometimes it happens that some extra characters are left in the input stream while taking the input such as, the ?\n? (Enter) character. This extra character is then passed to the next input and may pose problem.

To get rid of such extra characters the cin::ignore( ) function is used. This is equivalent to fflush ( stdin ) used in C language. This function ignores the first n characters (if present) in the input stream, stops if delimiter d is encountered.

 

17. Write a program that implements a date class containing day, month and year as data members. Implement assignment operator and copy constructor in this class.

Ans: This is shown in following program:

#include

class date

{

private :

int day ;

int month ;

int year ;

public :

date ( int d = 0, int m = 0, int y = 0 )

{

day = d ;

month = m ;

year = y ;

}

// copy constructor

date ( date &d )

{

day = d.day ;

month = d.month ;

year = d.year ;

}

// an overloaded assignment operator

date operator = ( date d )

{

day = d.day ;

month = d.month ;

year = d.year ;

return d ;

}

void display( )

{

cout << day << "/" << month << "/" << year ;

}

} ;

void main( )

{

date d1 ( 25, 9, 1979 ) ;

date d2 = d1 ;

date d3 ;

d3 = d2 ;

d3.display( ) ;

}

 

18. When should I use unitbuf flag?

Ans: The unit buffering (unitbuf) flag should be turned on when we want to ensure that each character is output as soon as it is inserted into an output stream. The same can be done using unbuffered output but unit buffering provides a better performance than the unbuffered output.

 

19.What are manipulators?

Ans: Manipulators are the instructions to the output stream to modify the output in various ways. The manipulators provide a clean and easy way for formatted output in comparison to the formatting flags of the ios class. When manipulators are used, the formatting instructions are inserted directly into the stream. Manipulators are of two types, those that take an argument and those that don?t.

 

20. What is the difference between the manipulator and setf( ) function?

Ans: The difference between the manipulator and setf( ) function are as follows:

 

The setf( ) function is used to set the flags of the ios but manipulators directly insert the formatting instructions into the stream. We can create user-defined manipulators but setf( ) function uses data members of ios class only. The flags put on through the setf( ) function can be put off through unsetf( ) function. Such flexibility is not available with manipulators.

 

21. How do I get the current position of the file pointer?

Ans: We can get the current position of the file pointer by using the tellp( ) member function of ostream class or tellg( ) member function of istream class. These functions return (in bytes) positions of put pointer and get pointer respectively.

 

22. What are put and get pointers?

Ans: These are the long integers associated with the streams. The value present in the put pointer specifies the byte number in the file from where next write would take place in the file. The get pointer specifies the byte number in the file from where the next reading should take place.

 

23. What do the nocreate and noreplace flag ensure when they are used for opening a file?

Ans: nocreate and noreplace are file-opening modes. A bit in the ios class defines these modes. The flag nocreate ensures that the file must exist before opening it. On the other hand the flag noreplace ensures that while opening a file for output it does not get overwritten with new one unless ate or app is set. When the app flag is set then whatever we write gets appended to the existing file. When ate flag is set we can start reading or writing at the end of existing file.

 

24. What is the limitation of cin while taking input for character array?

Ans: To understand this consider following statements,

char str[5] ;

cin >> str ;

While entering the value for str if we enter more than 5 characters then there is no provision in cin to check the array bounds. If the array overflows, it may be dangerous. This can be avoided by using get( ) function. For example, consider following statement,

cin.get ( str, 5 ) ;

On executing this statement if we enter more than 5 characters, then get( ) takes only first five characters and ignores rest of the characters. Some more variations of get( ) are available, such as shown below:

get ( ch ) ? Extracts one character only

get ( str, n ) ? Extracts up to n characters into str

get ( str, DELIM ) ? Extracts characters into array str until specified delimiter (such as '\n'). Leaves delimiting character in stream.

get ( str, n, DELIM ) ? Extracts characters into array str until n characters or DELIM character, leaving delimiting character in stream.

 

25. What is the purpose of istream class?

Ans: The istream class performs activities specific to input. It is derived from the ios class. The most commonly used member function of this class is the overloaded >> operator which can extract values of all basic types. We can extract even a string using this operator.

 

26. Would the following code work?

#include

void main( )

{

ostream o ;

o << "Dream. Then make it happen!" ;

}

Ans: No! This is because we cannot create an object of the ostream class since its constructor and copy constructor are declared private.

 

27. Can we use this pointer inside static member function?

Ans: No! The this pointer cannot be used inside a static member function. This is because a static member function is never called through an object.

 

28. What is strstream?

Ans: strstream is a type of input/output stream that works with the memory. It allows using section of the memory as a stream object. These streams provide the classes that can be used for storing the stream of bytes into memory. For example, we can store integers, floats and strings as a stream of bytes. There are several classes that implement this in-memory formatting. The class ostrstream derived from ostream is used when output is to be sent to memory, the class istrstream derived from istream is used when input is taken from memory and strstream class derived from iostream is used for memory objects that do both input and output. Ans: When we want to retrieve the streams of bytes from memory we can use istrestream. The following example shows the use of istrstream class.

#include

void main( )

{

int age ;

float salary ;

char name[50] ;

char str[] = "22 12004.50 K. Vishwanatth" ;

istrstream s ( str ) ;

s >> age >> salary >> name ;

cout << age << endl << salary << endl << name ;

cout << endl << s.rdbuf( ) ;

}

Here, s is the object of the class istrstream. When we are creating the object s, the constructor of istrstream gets called that receives a pointer to the zero terminated character array str. The statement s >> age >> salary >> name ; extracts the age, salary and the name from the istrstream object s. However, while extracting the name, only the first word of name gets extracted. The balance is extracted using rdbuf( ).

 

29. When the constructor of a base class calls a virtual function, why doesn't the override function of the derived class gets called?

Ans: While building an object of a derived class first the constructor of the base class and then the constructor of the derived class gets called. The object is said an immature object at the stage when the constructor of base class is called. This object will be called a matured object after the execution of the constructor of the derived class. Thus, if we call a virtual function when an object is still immature, obviously, the virtual function of the base class would get called. This is illustrated in the following example.

#include

class base

{

protected :

int i ;

public :

base ( int ii = 0 )

{

i = ii ;

show( ) ;

}

virtual void show( )

{

cout << "base's show( )" << endl ;

}

} ;

class derived : public base

{

private :

int j ;

public :

derived ( int ii, int jj = 0 ) : base ( ii )

{

j = jj ;

show( ) ;

}

void show( )

{

cout << "derived's show( )" << endl ;

}

} ;

 

void main( )

{

derived dobj ( 20, 5 ) ;

}

The output of this program would be:

base's show( )

derived's show( )

 

30. Can I have a reference as a data member of a class? If yes, then how do I initialise it?

Ans: Yes, we can have a reference as a data member of a class. A reference as a data member of a class is initialized in the initialization list of the constructor. This is shown in following program.

#include

class sample

{

private :

int& i ;

public :

sample ( int& ii ) : i ( ii )

{

}

void show( )

{

cout << i << endl ;

}

} ;

void main( )

{

int j = 10 ;

sample s ( j ) ;

s.show( ) ;

}

Here, i refers to a variable j allocated on the stack. A point to note here is that we cannot bind a reference to an object passed to the constructor as a value. If we do so, then the reference i would refer to the function parameter (i.e. parameter ii in the constructor), which would disappear as soon as the function returns, thereby creating a situation of dangling reference.

 

31. Why does the following code fail?

#include

class sample

{

private :

char *str ;

public :

sample ( char *s )

{

strcpy ( str, s ) ;

}

~sample( )

{

delete str ;

}

} ;

void main( )

{

sample s1 ( "abc" ) ;

}

Ans: Here, through the destructor we are trying to deal locate memory, which has been allocated statically. To remove an exception, add following statement to the constructor.

sample ( char *s )

{

str = new char[strlen(s) + 1] ;

strcpy ( str, s ) ;

}

Here, first we have allocated memory of required size, which then would get deal located through the destructor.

 

32. assert( ) macro...

We can use a macro called assert( ) to test for conditions that should not occur in a code. This macro expands to an if statement. If test evaluates to 0, assert prints an error message and calls abort to abort the program.

#include

#include

void main( )

{

int i ;

cout << "\nEnter an integer: " ;

cin >> i ;

assert ( i >= 0 ) ;

cout << i << endl ;

}

 

33. Why it is unsafe to deal locate the memory using free( ) if it has been allocated using new?

Ans: This can be explained with the following example:

#include

class sample

{

int *p ;

public :

sample( )

{

p = new int ;

}

~sample( )

{

delete p ;

}

} ;

void main( )

{

sample *s1 = new sample ;

free ( s1 ) ;

sample *s2 = ( sample * ) malloc ( sizeof ( sample ) ) ;

delete s2 ;

}

The new operator allocates memory and calls the constructor. In the constructor we have allocated memory on heap, which is pointed to by p. If we release the object using the free( ) function the object would die but the memory allocated in the constructor would leak. This is because free( ) being a C library function does not call the destructor where we have deal located the memory.

 

As against this, if we allocate memory by calling malloc( ) the constructor would not get called. Hence p holds a garbage address. Now if the memory is deal located using delete, the destructor would get called where we have tried to release the memory pointed to by p. Since p contains garbage this may result in a runtime error.

 

34. Can we distribute function templates and class templates in object libraries?

Ans: No! We can compile a function template or a class template into object code (.obj file). The code that contains a call to the function template or the code that creates an object from a class template can get compiled. This is because the compiler merely checks whether the call matches the declaration (in case of function template) and whether the object definition matches class declaration (in case of class template). Since the function template and the class template definitions are not found, the compiler leaves it to the linker to restore this. However, during linking, linker doesn't find the matching definitions for the function call or a matching definition for object creation. In short the expanded versions of templates are not found in the object library. Hence the linker reports error.

 

35. What is the difference between an inspector and a mutator ?

Ans: An inspector is a member function that returns information about an object's state (information stored in object's data members) without changing the object's state. A mutator is a member function that changes the state of an object. In the class Stack given below we have defined a mutator and an inspector.

class Stack

{

public :

int pop( ) ;

int getcount( ) ;

}

In the above example, the function pop( ) removes top element of stack thereby changing the state of an object. So, the function pop( ) is a mutator. The function getcount( ) is an inspector because it simply counts the number of elements in the stack without changing the stack.

 

36. Namespaces:

The C++ language provides a single global namespace. This can cause problems with global name clashes. For instance, consider these two C++ header files:

// file1.h

float f ( float, int ) ;

class sample { ... } ;

// file2.h

class sample { ... } ;

With these definitions, it is impossible to use both header files in a single program; the sample classes will clash. A namespace is a declarative region that attaches an additional identifier to any names declared inside it. The additional identifier thus avoids the possibility that a name will conflict with names declared elsewhere in the program. It is possible to use the same name in separate namespaces without conflict even if the names appear in the same translation unit. As long as they appear in separate namespaces, each name will be unique because of the addition of the namespace identifier. For example:

// file1.h

namespace file1

{

float f ( float, int ) ;

class sample { ... } ;

}

// file2.h

namespace file2

{

class sample { ... } ;

}

 

Now the class names will not clash because they become file1::sample and file2::sample, respectively.

 

37. What would be the output of the following program?

#include

class user

{

int i ;

float f ;

char c ;

public :

void displaydata( )

{

cout << endl << i << endl << f << endl << c ;

}

} ;

void main( )

{

cout << sizeof ( user ) ;

user u1 ;

cout << endl << sizeof ( u1 ) ;

u1.displaydata( ) ;

}

Ans: The output of this program would be,

9 or 7

9 or 7

Garbage

Garbage

Garbage

Since the user class contains three elements, int, float and char its size would be 9 bytes (int-4, float-4, char-1) under Windows and 7 bytes (int-2, float-4, char-1) under DOS. Second output is again the same because u1 is an object of the class user. Finally three garbage values are printed out because i, f and c are not initialized anywhere in the program.

 

Note that if you run this program you may not get the answer shown here. This is because packing is done for an object in memory to increase the access efficiency. For example, under DOS, the object would be aligned on a 2-byte boundary. As a result, the size of the object would be reported as 6 bytes. Unlike this, Windows being a 32-bit OS the object would be aligned on a 4-byte boundary. Hence the size of the object would be reported as 12 bytes. To force the alignment on a 1-byte boundary, write the following statement before the class declaration.

#pragma pack ( 1 )

 

38. Write a program that will convert an integer pointer to an integer and vice-versa.

Ans: The following program demonstrates this.

#include

void main( )

{

int i = 65000 ;

int *iptr = reinterpret_cast ( i ) ;

cout << endl << iptr ;

iptr++ ;

cout << endl << iptr ;

i = reinterpret_cast ( iptr ) ;

cout << endl << i ;

i++ ;

cout << endl << i ;

}

 

39. What is a const_cast?

Ans. The const_cast is used to convert a const to a non-const. This is shown in the following

program:

#include

void main( )

{

const int a = 0 ;

int *ptr = ( int * ) &a ; //one way

ptr = const_cast_ ( &a ) ; //better way

}

Here, the address of the const variable a is assigned to the pointer to a non-const variable. The const_cast is also used when we want to change the data members of a class inside the const member functions. The following code snippet shows this:

class sample

{

private:

int data;

public:

void func( ) const

{

(const_cast (this))->data = 70 ;

}

} ;

 

40. What is forward referencing and when should it be used?

Ans: Forward referencing is generally required when we make a class or a function as a friend.

Consider following program:

class test

{

public:

friend void fun ( sample, test ) ;

} ;

class sample

{

public:

friend void fun ( sample, test ) ;

} ;

void fun ( sample s, test t )

{

// code

}

void main( )

{

sample s ;

test t ;

fun ( s, t ) ;

}

On compiling this program it gives error on the following statement of test class. It gives an error that sample is undeclared identifier. friend void fun ( sample, test );

This is so because the class sample is defined below the class test and we are using it before its definition. To overcome this error we need to give forward reference of the class sample before the definition of class test. The following statement is the forward reference of class sample.

class sample ;

 

41. How would you give an alternate name to a namespace?

Ans: An alternate name given to namespace is called a namespace-alias. namespace-alias is generally used to save the typing effort when the names of namespaces are very long or complex. The following syntax is used to give an alias to a namespace.

namespace myname = my_old_very_long_name ;

 

42. Using a smart pointer can we iterate through a container?

Ans: Yes. A container is a collection of elements or objects. It helps to properly organize and store the data. Stacks, linked lists, arrays are examples of containers. Following program shows how to iterate through a container using a smart pointer.

#include

class smartpointer

{

private :

int *p ; // ordinary pointer

public :

smartpointer ( int n )

{

p = new int [ n ] ;

int *t = p ;

for ( int i = 0 ; i <= 9 ; i++ )

*t++ = i * i ;

}

int* operator ++ ( int )

{

return p++ ;

}

int operator * ( )

{

return *p ;

}

} ;

void main( )

{

smartpointer sp ( 10 ) ;

for ( int i = 0 ; i <= 9 ; i++ )

cout << *sp++ << endl ;

}

Here, sp is a smart pointer. When we say *sp, the operator * ( ) function gets called. It returns the integer being pointed to by p. When we say sp++ the operator ++ ( ) function gets called. It increments p to point to The next element in the array and then returns the address of this new location.

 

43. Can objects read and write themselves?

Ans: Yes! This can be explained with the help of following example:

#include

#include

class employee

{

private :

char name [ 20 ] ;

int age ;

float salary ;

public :

void getdata( )

{

cout << "Enter name, age and salary of employee : " ;

cin >> name >> age >> salary ;

}

void store( )

{

ofstream file ;

file.open ( "EMPLOYEE.DAT", ios::app | ios::binary ) ;

file.write ( ( char * ) this, sizeof ( *this ) ) ;

file.close( ) ;

}

void retrieve ( int n )

{

ifstream file ;

file.open ( "EMPLOYEE.DAT", ios:

 

 

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