Help with my printf function - c

Question

For debugging purposes I would like to have a printf_debug function that would function just like the standard printf function, but would only print if a #DEFINE DEBUG was true
I know I have to use varagrs (...) but I have no idea how to actually achieve that.
Thanks in advance.

Answer 1

Easier to just #define it away. Something like this:
#ifdef DEBUG
#define printf_debug printf
#else
#define printf_debug while(0)printf
#endif

Answer 2

I don't what exactly you want to achieve. In case you want a code block to execute only if DEBUG is defined, use the preprocessor directive #ifdef.
#include <stdio.h>
#include <stdarg.h>
#define DEBUG
void printf_debug(const char *format, ...) {
#ifdef DEBUG
va_list args;
va_start(args, format);
vprintf(format, args);
va_end(args);
#endif /* DEBUG */
}

Answer 3

You don't need to use vargs, macros will work. Here is an example, which will prints function and line number as well:
#ifdef DEBUG
#define printf_debug(fmt, args...) printf("%s[%d]: "fmt, __FUNCTION__, __LINE__, ##args)
#else
#define printf_debug(fmt, args...)
#endif
The ##args here will be replaced by the args list, which likes what vargs does in function call.

Answer 4

You have to use the va_arg macros, they are used to access the variadic variables. A useful link: http://www.cppreference.com/wiki/c/other/va_arg. The reference is for C++, but these macros can be used in C as well.
In your actual implementation you place the code using the variadic variables in a #ifdef block.
But if you're looking for a regular call to printf, dependent on DEBUG a simple #define acting as an alias will do.

Answer 5

C99 compiler only!
#include <stdio.h>
#define DEBUG
#ifdef DEBUG
#define debug(...) printf(__VA_ARGS__)
#else
#define debug while(0)
#endif
int main(int argc, char *argv[])
{
debug("Only shows when DEBUG is defined!\n");
return 0;
}
to be honest varidic macros are not needed you could just easily write it as this:
#include <stdio.h>
#define DEBUG
#ifdef DEBUG
#define debug printf
#else
#define debug while(0)
#endif
int main(int argc, char *argv[])
{
debug("Only shows when DEBUG is defined!\n");
return 0;
}
Thinking about it, debug information should go to stderr so as not to interfer with stdout, so this one should be favoured:
#include <stdio.h>
#define DEBUG
#ifdef DEBUG
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug while(0)
#endif
int main(int argc, char *argv[])
{
debug("Only shows when DEBUG is defined!\n");
return 0;
}