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19/07/07 - 361 Views - Ratings :   3 of 5 / 1 Votes

Level - Beginner
      

Question -

What will be  the output of the following code and why?

#include <stdio.h>

void f1(int *, int);   
void f2(int *, int);   
void(*p[2]) ( int *, int);  

int  main()  
{    
  int a;    
  int b;      
  p[0] = f1;   
  p[1] = f2;   
  a=3;    
  b=5;     
  p[0](&a , b);    
  printf("%d\t %d\t" , a ,b);     
  p[1](&a , b);   
  printf("%d\t %d\t" , a ,b); 
 }   

void f1( int* p , int q)  

{   
   int tmp;    
   tmp =*p;    
   *p = q;    
   q= tmp;  
}    

void f2( int* p , int q)  
{    
   int tmp;    
   tmp =*p;    
   *p = q;    
    q= tmp;  
}  

Answer -

 5 5 5 5

Explanation -

First thing to note is p in above c code is an array of pointer to a function.

Second thing to observe is f1 and f2 are exactly alike in all aspects.

Third thing to see is that a is passed by refrence and b is passed by value.

And finally since value of a is passed by refrence and that of b is passed by value, value of a can be modified to b while actual value of b cannot be modified to a. So all we get is 5 for both the variables every time.

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