#define for unsigned long

Question

Hi, I'm attempting to use the #define directive to change all of "ulong" to "unsigned long". Here is an example:

#define ulong unsigned long
ulong idCounter = 0;

Sadly, I think it ends up replacing ulong with "unsigned", rather than "unsigned long". I tried "#define ulong (unsigned long)", but that didn't worth either.

Answer 1

why don't you use just typedef?

typedef unsigned long ulong;

Answer 2

Your macro works OK:

cpp ulong.c

gives

# 1 "ulong.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "ulong.c"

unsigned long idCounter = 0;

I originally assumed that you already had a typedef called ulong and were trying to override that with the preprocessor, but other people have a different take on it. If you actually want the ulong, you should use a typedef, as they said.

Answer 3

Better use a typedef. The reason your macro may fail is - it might not syntactically valid in some places. Consider:

double x = calc();
ulong v = ulong(x);

In this case, you get

unsigned long v = unsigned long(x);

This is not valid, because the cast form used is not compatible with the way you name the type (it has to consist of a simplier form, like a single word). Use a typedef:

typedef unsigned long ulong;