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+21 votes

What does the following program print?


void f(int *p, int *q) {

int i=0, j=1;

int main() {
    f(&i, &j);
    printf("%d %d\n", i,j);
    return 0;
  1. $2 \ 2$
  2. $2 \ 1$
  3. $0 \ 1$
  4. $0 \ 2$
in Programming by Veteran (105k points)
edited by | 2.5k views
Why would someone pass address of global variable to a function??

2 Answers

+34 votes
Best answer
 p=q; // now p and q are pointing to same address i.e. address of j
 *p=2;// value of j will be updated to 2

Hence, answer is (D) 0 2

by Boss (44.1k points)
edited by
+8 votes

int  i=0, j=1;

The above statement declares and initializes two global variables.

void f(int *p ,int *q)
    % A %   p=q  // The address stored in pointer 'q' is assigned to pointer'p'
    % B %   *p=2  //The value at the address stored in pointer 'p' is set as 2. 

Now in the main function the function 'f' is called with the  address of variable 'i'   and 'j' as parameters.

Let us assume address of i=100 and address of j=200.

So the statement A in the definition of function f changes the value of the pointer 'p' from 100 to 200 as the pointer 'p' takes the address of pointer 'q'.

So after statement A the pointer 'p' also stores the address of j.

And in the statement B the value at address stored in the pointer 'p' is set to 2 this means the value of j is changed to 2.

Because of this the printf function in the main will print the value of i=0 and the value of j=2.

Therefore the output will be  0  2

by (385 points)
I have a doubt. In function "f" no value is being returned,then also value of J will get changed?
@Divyanka, we are sending address of the variable and modifying the value at that address which will definitely change the value of the variable.
ok. thank u

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