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Identify the error in the given C code snippet:

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#include <stdio.h>

int main() {
    int x = 5;
    int y = 10;
    swap(x, y);
    printf("After swapping: x = %d, y = %d\n", x, y);
    return 0;
}

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

Show answer & explanation

Answer: B. Incorrect use of pointers: swap expects pointer arguments (&x, &y), but integer values are passed

The swap function is declared to take two int* (pointer to int) parameters, but the function call swap(x, y) passes integer values directly instead of their addresses. The correct call should be swap(&x, &y) to pass pointers to the variables so they can be modified. Without this, the code will fail to compile with a type mismatch error.

Step-by-step Derivation:
Analysis of the code flow:

Function declaration: void swap(int *a, int *b) expects two pointers to integers.
Function call: swap(x, y) passes two integer values (not pointers).
Type mismatch: int cannot be implicitly converted to int*.
Compiler error: "incompatible type for argument" or similar.
Fix: Change the call to swap(&x, &y) to pass addresses using the address-of operator &.
Option A is incorrect: main() has a return statement (return 0).
Option C is incorrect: printf format specifiers %d correctly match int arguments.
Option D is incorrect: temp is properly declared as int and can hold integer values.

Identify the error in the given C code snippet:

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