Different output from macros and function definition

Question

#include<stdio.h>
#define prod(a,b) ((a>b)?a*a:b*b)

int prod1(int a,int b){
       return ((a>b)?a*a:b*b);
}

int main(){
        int p=0,q=-1;
        int p1=0,q1=-1;
        printf("%d ",prod(p1++,q1++));
        printf("%d ",prod1(p++,q++));
        return 0;
}

Output is: 1 0

Why is it different? How is the macros definition different from the function definition and why does it produces different results? Shouldn't all 2 outputs be equal to 0?

Modifying an unsequenced variable more than once is undefined behavior. — Captain Obvlious, Oct 01 '13 at 21:00
You have undefined behaviour in the first case (when using the macro). Expand the macro out manually and you'll see that you have unsequenced accesses and modifications to the `p` and `q` objects. — Joseph Mansfield, Oct 01 '13 at 21:00
With all due respect, I think we all have been bitten by this bug once. I know I have. It's seems obvious now, but it's subtle when you aren't thinking too much about the preprocessor. The #define min/max version is a more common one. — Charlie Burns, Oct 01 '13 at 23:42

score 3 · Accepted Answer · answered Oct 01 '13 at 21:01

3

In the macro

#define prod(a,b) ((a>b)?a*a:b*b)

a and b are referenced more than once. So

prod(p1++,q1++)

becomes

(p1++ > q1++) ? p1++*p1++ : q1++*q2++

Note that this is very different than calling the function

prod1(p1++, q1++)

which only increments p1 and q1 once. And does so predictably.

answered Oct 01 '13 at 21:01

Charlie Burns

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20
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Oh, right. Should have expanded it first before asking. My bad. – Abhinav Singi Oct 01 '13 at 21:09
@haccks No actually should have expanded it manually, then wouldn't have even searched it on SO. – Abhinav Singi Oct 01 '13 at 21:15

markgz · Answer 2 · 2013-10-01T21:08:49.993

0

The macro's arguments a and b are copied as text into the expanded body of the macro, so prod(p1++,q1++) expands to ((p1++>q1++)?p1++*p1++:q1++*q1++) which is obviously different to what the function evaluates.

In standard C it is undefined if a single object is modified more than once between successive sequence points, so the macro causes undefined behaviour.

edited Oct 01 '13 at 21:08

answered Oct 01 '13 at 21:02

markgz

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1
19
41

score 0 · Answer 3 · answered Oct 01 '13 at 21:03

Look for the code that is compiled after the pre-processor has finished its jobs:

int prod1(int a,int b){
       return ((a>b)?a*a:b*b);
}

int main(){
    int p=0,q=-1;
    int p1=0,q1=-1;
    int p2=0,q2=-1;
    printf("%d ",((p1++>q1++)?p1++*p1++:q1++*q1++));
    printf("%d ",prod1(p++,q++));
    printf("%d ",(p2>q2)?p2*p2:q2*q2);
    return 0;
}

Different output from macros and function definition

3 Answers3