Suppose you have
#define mul(x, y) x * y
What happens if I say:
mul(a + 5, 6); /* a + 5 * 6 */
Now if I slighlty change the macro:
#define mul(x, y) ((x) * (y))
mul(a + 5, 6); /* ((a + 5) * (6)) */
Remember, the arguments aren't evaluated or anything, only textual substitution is performed.
EDIT
For an explanation about having the entire macro in parentheses, see the link posted by Nate C-K.
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