How to wrap printf() into a function or macro?

Since you can use C99, I'd wrap it in a variadic macro:

#define TM_PRINTF(f_, ...) printf((f_), __VA_ARGS__)
#define TM_SNPRINTF(s_, sz_, f_, ...) snprintf((s_), (sz_), (f_), __VA_ARGS__)

since you didn't say that you have vprintf or something like it. If you do have something like it, you could wrap it in a function like Sergey L has provided in his answer.

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The above TM_PRINTF does not work with an empty VA_ARGS list. At least in GCC it is possible to write:

#define TM_PRINTF(f_, ...) printf((f_), ##__VA_ARGS__)

The two ## signs remove the excess comma in front of them them if __VA_ARGS__ is empty.

If you can live with having to wrap the call in two parentheses, you can do it like this:

#define THAT_MACRO(pargs)    printf pargs

Then use it:

THAT_MACRO(("This is a string: %s\n", "foo"));
           ^
           |
          OMG

This works since from the preprocessor's point of view, the entire list of arguments becomes one macro argument, which is substituted with the parenthesis.

This is better than just plain doing

#define THAT_MACRO printf

Since it allows you to define it out:

#define THAT_MACRO(pargs)  /* nothing */

This will "eat up" the macro arguments, they will never be part of the compiled code.

UPDATE Of course in C99 this technique is obsolete, just use a variadic macro and be happy.

There are 2 ways to do this:

1. Variadric macro

   #define my_printf(...) printf(__VA_ARGS__)
   

2. function that forwards va_args

   #include <stdarg.h>
   #include <stdio.h>
   
   void my_printf(const char *fmt, ...) {
       va_list args;
       va_start(args, fmt);
       vprintf(fmt, args);
       va_end(args);
   }
   

There are also vsnprintf, vfprintf and whatever you can think of in stdio.