How can I use #if inside #define in the C preprocessor?

Question

I want to write a macro that spits out code based on the Boolean value of its parameter. So say DEF_CONST(true) should be expanded into const, and DEF_CONST(false) should be expanded into nothing.

Clearly the following doesn't work because we can't use another preprocessor inside #defines:

#define DEF_CONST(b_const) \
#if (b_const) \
  const \
#endif

What is the point of this? Do you have a concrete example, because this seems rather strange... — Dean Harding
– Dean Harding, Commented May 14, 2010 at 5:06
See also <stackoverflow.com/questions/8427963/if-inside-define>. — Rhubbarb
– Rhubbarb, Commented Dec 8, 2011 at 9:31
@DeanHarding Didn't exist at the time of your comment, but _Generic is an example. You may need to check if uintmax_t has a different underlying type than unsigned long long. — 12431234123412341234123
– 12431234123412341234123, Commented Jul 8, 2021 at 15:19

vladr · Accepted Answer · 2011-09-27 13:46:45Z

70

You can simulate conditionals using macro token concatenation as follows:

#define DEF_CONST(b_const) DEF_CONST_##b_const
#define DEF_CONST_true const
#define DEF_CONST_false

Then,

/* OK */
DEF_CONST(true)  int x;  /* expands to const int x */
DEF_CONST(false) int y;  /* expands to int y */

/* NOT OK */
bool bSomeBool = true;       // technically not C :)
DEF_CONST(bSomeBool) int z;  /* error: preprocessor does not know the value
                                of bSomeBool */

Also, allowing for passing macro parameters to DEF_CONST itself (as correctly pointed out by GMan and others):

#define DEF_CONST2(b_const) DEF_CONST_##b_const
#define DEF_CONST(b_const) DEF_CONST2(b_const)
#define DEF_CONST_true const
#define DEF_CONST_false

#define b true
#define c false

/* OK */
DEF_CONST(b) int x;     /* expands to const int x */
DEF_CONST(c) int y;     /* expands to int y */
DEF_CONST(true) int z;  /* expands to const int z */

You may also consider the much simpler (though potentially less flexible):

#if b_const
# define DEF_CONST const
#else /*b_const*/
# define DEF_CONST
#endif /*b_const*/

edited Sep 27, 2011 at 13:46

user142162

answered May 14, 2010 at 5:11

vladr

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3 Comments

GManNickG Over a year ago

Typo in the above, just missed the edit mark. Should be #define DEF_CONST(b_const) CONCATENATE(DEF_CONST_, b_const).

Michael Burr Over a year ago

Yup - token pasting or stringizing ('##' or '#') preprocessor operators almost always really need a level of indirection to work as desired: stackoverflow.com/questions/1767683/…

tdihp Over a year ago

Possible to expand this so it takes parameter as one would use in x macro?

Peter Mortensen · Accepted Answer · 2025-03-23 23:30:45Z

8

Doing it as a parametrised macro is a bit odd.

Just do something like this:

#ifdef USE_CONST
    #define MYCONST const
#else
    #define MYCONST
#endif

Then you can write code like this:

MYCONST int x = 1;
MYCONST char* foo = "bar";

and if you compile with USE_CONST defined (e.g., typically something -DUSE_CONST in the makefile or compiler options) then it will use the consts; otherwise it won't.

edited Mar 23 at 23:30

Peter Mortensen

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answered May 14, 2010 at 9:08

GrahamS

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