Does the sizeof operator work in preprocessor #if directives?

Question

Can we use the sizeof operator in #if macros? If yes, how? And if not, why?

Does the sizeof operator work in preprocessor #if directives?

Answer 1

No; the sizeof() operator does not work in C preprocessor conditional directives such as #if and #elif.

The reason is that the C pre-processor does not know a thing about the sizes of types.

You can use sizeof() in the body of a #define'd macro, of course, because the compiler handles the analysis of the replacement text and the preprocessor does not. For example, a classic macro gives the number of elements in an array; it goes by various names, but is:

#define DIM(x) (sizeof(x)/sizeof(*(x)))

This can be used with:

static const char *list[] =
{
    "...", ...
};
size_t size_list = DIM(list);

What you can't do is:

#if sizeof(long) > sizeof(int)  // Invalid, non-working code
...
#endif

(The trouble is that the condition is evaluated to #if 0(0) > 0(0) under all plausible circumstances, and the parentheses make the expression invalid, even under the liberal rules of preprocessor arithmetic.)

Answer 2

A sizeof() operator cannot be used in #if and #elif line because the preprocessor does not parse type names.

But the expression in #define is not evaluated by the preprocessor; hence it is legal in #define case.