#define func(x, y) x + y/x

Question

#include <stdio.h>
#define func(x, y) x + y/x
int main() {
    int i = -1, j = 2, x;
    printf("i = %d\n", i);
    printf("j = %d\n", j);
    printf("x = %d\n", x);
    x = func(i + j, 3);
    printf("%d\n",x);
    return 0;
}

In the C code above, the output is 0 while I expect 4. i.e.

i + j = -1 + 2 = 1   
func(i+j, 3) = func (1,3) = 1 + 3/1 = 1 + 3 = 4.

Where am I wrong? Where will I get to learn more about C-pre-processor macro behavior?

The output of the above code is as below:

https://gcc.gnu.org/onlinedocs/cpp/Macro-Pitfalls.html You have couple of these. — Eugene Sh., Aug 03 '17 at 16:11
[The need for parentheses in macros in C](https://stackoverflow.com/q/10820340/995714), and using `x` before initializing is [undefined behavior](https://stackoverflow.com/q/11962457/995714) — phuclv, Aug 03 '17 at 16:22
Run `cpp` on your source file and look at the resulting code. It will be obviouss. — lurker, Aug 03 '17 at 16:29
Tip: for future posts, include text as text, rather than pictures of text. — chux - Reinstate Monica, Aug 03 '17 at 16:45

skr · Accepted Answer · 2017-08-03T16:34:09.657

3

change

#define func(x, y) x + y/x

To

#define func(x, y) ((x) + (y)/(x))

Reason :

func(x, y) x + y/x  
x = i + j
y = 3
func(x, y) = func(i + j, 3)
=  x + y/x
=  i + j + 3/i + j
= -1 + 2 + 3/(-1) + 2
=  1 - 3 + 2
=  0

edited Aug 03 '17 at 16:34

answered Aug 03 '17 at 16:30

skr

2,146
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22

4

Or better still: `#define func(x, y) ((x) + (y)/(x))` – Paul R Aug 03 '17 at 16:31
2

now call `func(1, 2+3)` or `func(1, 2+3)*4` and see – phuclv Aug 03 '17 at 16:33
Sorry about that. Thank you guys. – skr Aug 03 '17 at 16:35

#define func(x, y) x + y/x

1 Answers1