GATE CS 2017 Set 2 Q.14 (CS Q.14): Pointers, Arrays and Strings

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

0, 0

0, 1

1, 0

1, 1

Official answer and solution

Solution

Trace the pointer assignments and memory modifications step by step.

Trace the execution of printxy(1, 1):

Parameters x and y are initialized to 1 (local copies passed by value).
int *ptr;: Declares a local pointer variable ptr.
x = 0;: The local variable x becomes 0.
ptr = &x;: The pointer ptr now stores the address of local variable x.
y = *ptr;: Dereferences ptr (value of x, which is 0) and assigns it to local variable y. Thus, y becomes 0.
*ptr = 1;: Modifies the memory location pointed to by ptr (which is x) to 1. Thus, x becomes 1.
printf("%d, %d", x, y);: Prints the current values of x (1) and y (0).

The printed output is 1, 0, corresponding to option C.

Final answer

Common trap

Assuming *ptr = 1 changes y because y had received *ptr earlier; y = *ptr copied the value, not the reference.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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