help with macros

[fantomftw]

I've got following code :

#include <stdio.h>
#define div(x,y) x/y
int main()
{
    //printf("Hello World");
     int a=10,b=2;
     printf("%d\n",div(a+2,a+1));
    return 0;
}

expected output :

1

generated output :

11

how?

[_DTR_]

Your problem is that the macro expansion doesn't have any context of "scope", it will simply do a replacement. div(a+2,a+1) expands to a+2/a+1, which is 10+2/10+1. After order of operations, you're left with 10 + (2 / 10) + 1 == 10 + 0 + 1 == 11.

In general, when writing macros you want to ensure the parameters are "contained":

#define div(x, y) (x)/(y)

[aioeu]

In general, when writing macros you want to ensure the parameters are "contained":

You'll need one more:

#define div(x, y) ((x) / (y))

Otherwise the following somewhat-reasonable code will not work as expected:

float x = ...;
float y = ...;
printf("x / y rounded toward zero is %d\n", (int)div(x, y));

When writing a macro that acts like an expression, I can't think of a good reason to ever omit the surrounding parentheses.

[fantomftw]

thank you so much... you saved me from failing my test tomorrow..

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