Question

Given the following function definition,

void shift(int*a, int*b)
(
*a = b
*b = *a;
}
what is the output after the following function call
int first = 0, second = 10;
shift(7first, &second);
printf('%d %d \\"", first, secon);
A) 10 0
B) 0 0
C) 10 10
D) 0 10

Answer 1

The C function 'shift' contains an error that results in both variables having the value of 'second' after the function is called. The correct output is '10 10'.

Step-by-step explanation:

The question provided relates to the execution of a C function and its side effects. The function shift is meant to swap the values of two integer variables.

However, there is an error in the function's implementation that does not achieve the intended swap. The line *a = *b; assigns the value of *b to *a, but the subsequent line *b = *a; assigns the value that *a just received (which is the original value of *b) back to *b, effectively not changing the value of *b.

Thus, after calling shift(&first, &second);, and then executing printf('%d %d \\', first, second);, the output will be 10 10.