GATE CSE 2017 | Set 2 | Question: 14

Question

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:

• A. $0, 0$
• B. $0, 1$
• C. $1, 0$
• D. $1, 1$

Comments

C

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Here is the solution : 

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This is independent of input arguments.

Answer 1

At first in loop we are giving $x=0$ then $ptr$ is pointing to $X$.

So, $*ptr=0$

Now, we copying the value of $ptr$ to $y$ ,so $Y=0$

 x=0;      //value of x = 0  
    ptr= &x;      // ptr points to variable x
    y= *ptr;      // Y contain value pointed by ptr i.e. x= 0;

Now, value of $ptr$ is changed to $1$. so the location of $X$ itself got modified

 *ptr=1;  

As it is pointing to $x$ so $x$ will also be changed to $1$

So, $1,0$ will be the value

C is correct answer here.

Answer 2

Option c is right.

Answer 3

we got x= 1 and y= 1;
void printxy(int x, int y) 
{  
    int *ptr;     //pointer is created which contain integer value.
    x=0;          //value of x = 0 here. 
    ptr= &x;      // ptr point to variable which has 0 
    y= *ptr;      // y contain value pointed by ptr i.e. x= 0;
    *ptr=1;       // value pointer by ptr is now set to 1 i.e. x= 1;
    printf("%d,%d",x,y);  // print x,y = x= 1 y=0

}

C is answer

Answer 4

hope it might help.......

Answer 5

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 6

C is the answer