recategorized by
9,550 views
30 votes
30 votes

Consider the following function implemented in C:

void printxy(int x, int y) {
    int *ptr;
    x=0;
    ptr=&x;
    y=*ptr;
    *ptr=1;
    printf(“%d, %d”, x, y);
}

The output of invoking $printxy(1,1)$ is:

  1. $0, 0$
  2. $0, 1$
  3. $1, 0$
  4. $1, 1$
recategorized by

11 Answers

2 votes
2 votes

Here's the code with a full explanation.

void printxy(int x, int y) {
    int *ptr; 
    x=0;     /*The value of x is now 0.*/
    
    ptr=&x;  /*Assigning the address of x to ptr.
             It means ptr points to the address of x.
             So *ptr=0 as x = 0.
             */
            
    y=*ptr;  /*Assigning the value that the ptr holds, to y
              y=0 as *ptr=0.
             */
             
    *ptr=1;  /*Assigning 1 to the value that the ptr holds.
             It means x=*ptr=1 as ptr points to the address of x.
             So the value of x is now 1.
             */
             
             /*Therefore, x=1 and y=0 now */
             
    printf(“%d, %d”, x, y);  /*It will give the output as 1, 0 */
}

 

Note that the function will print the same value no matter what the parameters are passed into it.

So the correct answer is C.

Answer:

Related questions

70 votes
70 votes
13 answers
1
Madhav asked Feb 14, 2017
27,483 views
Consider the following C program.#include<stdio.h #include<string.h int main() { char* c="GATECSIT2017"; char* p=c; printf("%d", (int)strlen(c+2[p]-6[p]-1)); return 0; }T...
25 votes
25 votes
7 answers
3
Madhav asked Feb 14, 2017
11,774 views
Consider the following C program.#include<stdio.h int main () { int m=10; int n, n1; n=++m; n1=m++; n ; n1; n-=n1; printf(“%d”, n); return 0; }The output of the prog...
26 votes
26 votes
3 answers
4
khushtak asked Feb 14, 2017
5,770 views
Match the following:$$\begin{array}{|ll|ll|}\hline P. & \text{static char var ;} & \text{i.} & \text{Sequence of memory locations to store addresses} \\\hline Q. & \text...