GATE CSE 2015 | Set 1 | Question: 1‏1

Question

The output of the following C program is_____________.

void f1 ( int a, int b)  {  
                int c; 
                c = a; a = b;
                 b = c;  
}   
void f2 ( int * a, int * b) {   
               int c; 
               c = * a; *a = *b; *b = c; 
} 
int main () { 
        int a = 4, b = 5, c = 6; 
        f1 ( a, b); 
        f2 (&b, &c); 
        printf ("%d", c - a - b);  
 }

Comments

Abb itne easy question kyo nhi aate?

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@Gok Sab solve karlenge to examiner ke question banane ka kya faayda :)

Answer 1

void f1 ( int a, int b)  {    //This code is call by value
// hence no effect of actual values when run.
                int c; 
                c = a; 
                a = b;
                b = c;  
}   
void f2 ( int * a, int * b) {   //*a= address of b 
//and *b = address of c
               int c;            //int c = garbage 
               c = * a;          //c = value at address a = 5;
               *a = *b;          //*a = Exchange original
// variable value of c to b = b= 6
               *b = c;             //*b = c = 5;
} 
int main () { 
        int a = 4, b = 5, c = 6; 
        f1 ( a, b);  This has no effect on actual values
// of a ,b since Call by value.
        f2 (&b, &c); Here change will be happen.
        At this point  int a = 4, b = 6, c = 5;
        printf ("%d", c - a - b);    = (5-4)-6 = 1-6 = -5
 }

Comments

passed as reference so there is change only in the second function

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can we give negative answers lik(-5) in gate NAT?

 

Answer 2

Here, $f1$ will not change any values bcz it is call by value but $f2$ is call by reference and it swaps values of $b$ and $c$ and changes are also reflected in main function. So, $5-4-6= -5$ is the answer.

Comments

I think there is a typing mistake by gate2015, it should be 5-4-6=-5 rather than 5-6-4=-5.

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the given program  f2 is call by referance

f2 does not support call by reference in c.

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$f2$ is not “call by reference”, $\text{C}$ only supports call by value. $f2$ is call by value, the value being the address of variable $b$ and $c.

Using pointers it mimics the behaviour of call by reference.

Answer 3

Here is the solution what I think would be correct but hey I may be wrong ..

Comments

answer is -5

Answer 4

Function $f1$ is for swapping but as parameters are passed by values so swapping of values will be done on former parameter (parameter local to $f1$) only not on actual parameter. So basically $f1$ is doing nothing here.

Function $f2$ is swapping values of $\text{b & c }$. As parameter are passed by reference ( addresses of actual variables are passed as parameter) so changes will reflect in actual variables also.

Answer 5

So answer is -5.