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Q:

Can you suggest any other way of writing the following expression such that 30 is used only once?

a <= 20 ? b = 30 : c = 30 ;

Answer

*( ( a <= 20 ) ? &b : &c ) = 30;

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Subject: Programming

Q:

What would be the output of the following program?

main()

{

  int i=2 ;

  printf ("\n%d%d", ++i, ++i );

}

Answer

Output may vary from compiler to compiler.


The order of evaluation of the arguments to a function call is unspecified.

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Subject: Programming

Q:

What would be the output of the following program?

main()

{

    static int a[20];

    int i = 0;

    a[i] = i++;

    printf ("\n%d%d%d", a[0],  a[1], i);

}

Answer

0  0  1


That's what some of the compilers would give. But some other compiler may give a different answer. The reason is, when a single expression causes the same object to be modified and then inspected the behaviour is undefined.

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Subject: Programming

Q:

Rewrite the following set of statements using conditional operators.

int a =1, b ;

if ( a > 10 )

b = 20; 

Answer

int a = 1, b , dummy;


a > 10 ? b = 20 : dummy =1;


 


Note that the following would not have worked:


a > 10 ? b = 20 : ;;


 

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Subject: Programming

Q:

Point out the error, if any, in the following program.

main()

{

    int i = 4, j = 2;

    switch(i)

    {

     case 1 :

       printf (''\n To err is human, to forgive is against company policy.");

        break;

      case j :

       printf (''\n if you have nothing to do, don't do it here.");

       break;

    }

}

Answer

Constant expression required in the second case, we cannot use j.

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Subject: Programming

Q:

Point out the error, if any, in the while loop.

main()

{

  int i = 1;

  while ()

  {

     printf ( "%d", i++);

     if (i >10) 

     break ;

   }

}

Answer

The condition in the while loop is a must.

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Subject: Programming

Q:

Point out the error, if any, in the following program.

main()

{

    int ( *p )() = fun;

    ( *P ) ();

}

fun ()

{

    Printf ( "\nLoud and clear" );

} 

Answer

Here we are initalising the function pointer p to the address of the function fun(). But during this initialisation the function has not been defined. Hence an error.


To eliminate this error add the prototype of the fun() before declaration of p, as shown below:


extern int fun();    or simply  int fun();

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Subject: Programming

Q:

What would be the output of the following program?

main()

{

    extern int fun ( float );

    int a;

    a = fun ( 3. 14 );

    printf ("%d", a);

}

int fun ( aa )

float aa ;

{

     return ( (int) aa );

}

Answer

Error occurs because we have mixed the ANSI prototype with K & R style of function definition.


When we use ANSI prototype for a function and pass a float to the function it is promoted to a double. When the function accepts this double into a float a type mismatch occurs hence the error.


The remedy for this error could be to define the function as :


int fun (float aa)


{


  ....


}

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Subject: Programming