Why macro function isn't call by value?

Question

#include<stdio.h>
#define exch(A,B) {int t=A;A=B;B=t;}
int main(void)
{
    int a=1,b=4;
    exch(a,b);
    printf("%d\t%d\n",a,b);
    return 0;
}

Prints:a=4,b=1

In the output, a and b have been exchanged; why is a macro not call by value? Please help me by explaining the result.

score 6 · Accepted Answer · answered Mar 01 '14 at 01:35

Because macros are not functions, the preprocessor just replaces the text. After the macro substitution, your program turns into:

//code in stdio.h

int main(void)
{
    int a=1,b=4;
    {int t=A;A=B;B=t;}
    printf("%d\t%d\n",a,b);
    return 0;
}

See? There's no function calls in the swapping.

1 Answers1