Using macro results in incorrect output when used as part of a larger math expression - why does this happen?

Question

This is a normal C routine program which i found out in some question bank. It is shown below:

#define CUBE(p) p*p*p

main()
{
    int k;
    k = 27 / CUBE(3         


        

Answer 1

C macros do textual substitution (i.e. it's equivalent to copying and pasting code). So your code goes from:

k=27/CUBE(3);

to

k=27/3*3*3;

Division and multiplication have the same precedence and have left-to-right associativity, so this is parsed as:

k=((27/3)*3)*3;

which is 9 * 3 * 3 = 81.

This is why C macros should always be defined with liberal use of parentheses:

#define CUBE(p) ((p) * (p) * (p))

For more information, see http://c-faq.com/cpp/safemacros.html from the comp.lang.c FAQ.