c - Why do these swap functions behave differently?

Question

#include <stdio.h>

void swap1(int a, int b)
{
    int temp = a;

    a = b;
    b = temp;
}

void swap2(int *a, int *b)
{
    int *temp = a;

    a = b;
    b = temp;
}

void swap3(int *a, int *b)
{
    int temp = *a;

    *a = *b;
    *b = temp;
}

main()
{
    int a = 9, b = 4;

    printf("%d , %d
", a, b);
    swap1(a, b);
    printf("%d , %d
", a, b);
    swap2(&a, &b);
    printf("%d , %d
", a, b);
    swap3(&a, &b);
    printf("%d , %d
", a, b);

}

Answer 1

C has value semantics for function parameters. This means the a and b for all your three swap variants are local variables of the respective functions. They are copies of the values you pass as arguments. In other words:

• swap1 exchanges values of two local integer variables - no visible effect outside the function
• swap2 exchanges values of two local variables, which are pointers in this case, - same, no visible effect
• swap3 finally gets it right and exchanges the values pointed to by local pointer variables.