#define macro for debug printing in C?

Trying to create a macro which can be used for print debug messages when DEBUG is defined, like the following pseudo code:

#define DEBUG 1
#define debug_print(args ...) if (DEBUG) fprintf(stderr, args)

How is this accomplished with a macro?

#define debug_print(FMT, ARGS...) do { \
if (DEBUG) \
fprintf(stderr, "%s:%d " FMT "\n", __FUNCTION__, __LINE__, ## ARGS); \
} while (0)

I would do something like

#ifdef DEBUG
#define debug_print(fmt, ...) fprintf(stderr, fmt, __VA_ARGS__)
#else
#define debug_print(fmt, ...) do {} while (0)
#endif


I think this is cleaner.

Here's the version I use:

#ifdef NDEBUG
#define Dprintf(FORMAT, ...) ((void)0)
#define Dputs(MSG) ((void)0)
#else
#define Dprintf(FORMAT, ...) \
fprintf(stderr, "%s() in %s, line %i: " FORMAT "\n", \
__func__, __FILE__, __LINE__, __VA_ARGS__)
#define Dputs(MSG) Dprintf("%s", MSG)
#endif

For a portable (ISO C90) implementation, you could use double parentheses, like this;

#include <stdio.h>
#include <stdarg.h>

#ifndef NDEBUG
# define debug_print(msg) stderr_printf msg
#else
# define debug_print(msg) (void)0
#endif

void
stderr_printf(const char *fmt, ...)
{
va_list ap;
va_start(ap, fmt);
vfprintf(stderr, fmt, ap);
va_end(ap);
}

int
main(int argc, char *argv[])
{
debug_print(("argv[0] is %s, argc is %d\n", argv[0], argc));
return 0;
}

or (hackish, wouldn't recommend it)

#include <stdio.h>

#define _ ,
#ifndef NDEBUG
# define debug_print(msg) fprintf(stderr, msg)
#else
# define debug_print(msg) (void)0
#endif

int
main(int argc, char *argv[])
{
debug_print("argv[0] is %s, argc is %d"_ argv[0] _ argc);
return 0;
}

According to

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,
there should be a `##` before `__VA_ARGS__`.

Otherwise, a macro `#define dbg_print(format, ...) printf(format, __VA_ARGS__)` will not compile the following example: `dbg_print("hello world");`.