Macro issue and what may be the output?

Question

#define CUBE(x)(x*x*x)
int main()
{
    int a, b;
    b = 3;
    a = CUBE(b++)/b++;
    printf(a=%d b=%d\n", a, b);
    return 0;
}

I have a confusion in this Macro defining statement? and I need output too?

Routine old undefined behaviour...what's the canonical question that this is a duplicate of? You can't modify `b` four times between sequence points and expect to get anything sane. Strictly, you can't invoke `printf()` without a prototype in scope either – you don't show the `#include ` line. However, that usually works. The `(b++*b++*b++)/b++` nonsense has no answer that you can expect — you can find what a specific compiler does, but it's under no obligation to do anything sane. — Jonathan Leffler, Sep 14 '13 at 05:20

score 1 · Answer 1 · answered Sep 14 '13 at 05:19

1

The part

CUBE(b++)

will be converted to

(b++ * b++ * b++)

which is undefined behavior or problematic due to modifying a variable in one single statement. It's recommended to avoid such a thing.

Try to pass a variable without ++ or --.

answered Sep 14 '13 at 05:19

masoud

score 0 · Answer 2 · answered Sep 14 '13 at 05:25

a=CUBE(b++)/b++;  

      |  
      V  


a=(b++ * b++ * b++)/b++ ;

In the above expression b value modifies between sequence points would cause undefined behavior due to lack of Sequence_point

To avoid this first assign and then increment

   a=CUBE(b)/b;
   b++;

score 0 · Answer 3 · answered Sep 14 '13 at 05:27

Use #define CUBE(x) ((x)*(x)*(x)) to avoid mistake when x is an expression.
a = CUBE(b++)/(b++); The value of a after executing the statement depends on the compiler you use or something else. This is called undefined behavior.

3 Answers3