Macro preprocessor directive

Question

I have idea about macro preprocessor directive.In below question how replacement in source code is done by compiler that it is giving output as "y is 392"

#include <stdio.h>
#define CUBE(x) (x*x*x)
void main(void)
{
     int x;
     int y;
     x = 5;
     y = CUBE(++x);
     printf("y is %d\n", y);
}

output:y is 392.

It's UB — undefined behaviour — because you end up with `y = (++x*++x*++x);` and you can't do multiple increments on a single variable between two sequence points (within a single assignment). Also, note too that the macro miscomputes the cube if used with `y = CUBE(x + 3);`. Use parentheses around arguments and around the expression — `#define CUBE(x) ((x)*(x)*(x))`. — Jonathan Leffler, Feb 06 '18 at 15:18
It's an undefined operation. The macro expands to `(++x*++x*++x)`. — Mad Physicist, Feb 06 '18 at 15:20

score 1 · Answer 1 · answered Feb 06 '18 at 15:18

1

The behaviour of your code is undefined.

CUBE(++x) expands to (++x * ++x * ++x): there are no sequencing points in that expression and it simultaneously reads and writes to x.

This epitomises why macros that do arithmetic are computationally lethal.

answered Feb 06 '18 at 15:18

Bathsheba

231,907
34
361
483

score 0 · Answer 2 · edited Feb 06 '18 at 15:20

0

The preprocessor is replacing that with

y = (++x*++x*++x);

Which is undefined behaviour.

edited Feb 06 '18 at 15:20

Jonathan Leffler

730,956
141
904
1,278

answered Feb 06 '18 at 15:19

Colin

3,394
1
21
29

Macro preprocessor directive

2 Answers2