C compiling error: stray '##' in program

Question

I was working with an embedded kernel source when I saw something like this:

#define OMAP_SYS_TIMER_INIT(name, clkev_nr, clkev_src, clksrc_nr, clksrc_src) \
static void __init omap##name##_timer_init(void)                              \
{                                                                             \
    omap2_gp_clockevent_init((clkev_nr), clkev_src);                          \
    omap2_gp_clocksource_init((clksrc_nr), clksrc_src);                       \
}

and when I tryed to make a program that used this ## thing (that I don't know the name) to see what it could really do I didn't got it to work. Below is what I did to test it's function. I just want to see if the argument inside the ## is literal or not, but something is clearly missing in my code for it to compile...

#include <stdio.h>
#include <stdlib.h>

#define DEFINE_1 2
#define DEFINE_2 4
#define DEFINE_3 6

#define DEFINE_i 9

int main(void)
{
  int i;
  for(i=1;i<4;i++) {
    printf("numero %d = %d\n",i,DEFINE_##i##);
  }
  return EXIT_SUCCESS;
}

The output of gcc is:

test.c: In function ‘main’:
test.c:14:5: error: stray ‘##’ in program
test.c:14:33: error: ‘DEFINE_’ undeclared (first use in this function)
test.c:14:33: note: each undeclared identifier is reported only once for each function it appears in
test.c:14:42: error: expected ‘)’ before ‘i’
test.c:14:42: error: stray ‘##’ in program

Anyone knows what is wrong? Thanks

Answer 1

It's the token concatenation operator for the C preprocessor. The reason your example doesn't compile is because you're not using the ## operator within a macro (i.e. #define statement).

Here's another post with some more information.

Answer 2

## means concatenation at time of preprocessing. http://gcc.gnu.org/onlinedocs/cpp/Concatenation.html