C preprocessor at run time?

I want to do a Token concatenation, but I want to do this with the content of a variable, not its name. like this.

#define call_function(fun, number) fun##_##number ()


while (i < 10 ) {
  call_function(fun, i);
 }

but I give fun_number (), I want to give fun_1, fun_2, and so on...

how to do it?

About function pointers.

I'm walking to handle the user input on ascii caracters, my option is & * ^ > < and so on up to ten tokens.

to handle it in a开发者_JAVA百科n array with function pointer, I need to create an array with 127 blocks for only ten tokens!

this is too expensive! thanks for all answers.

Thanks in advance!!!

#define call_function(fun, member) fun##_##number ()
// ------------------------^
// should be "number".

Even so, you'll only get fun_i. You can't call the preprocessor macros at runtime because they're used only in preprocessing (even before parsing and compiling).

You need to expand the loop manually.

call_function(fun, 0);
call_function(fun, 1);
call_function(fun, 2);
call_function(fun, 3);
call_function(fun, 4);
call_function(fun, 5);
call_function(fun, 6);
call_function(fun, 7);
call_function(fun, 8);
call_function(fun, 9);

Or use __COUNTER__ (needs gcc ≥ 4.3):

#define CONCAT3p(x,y,z) x##y##z
#define CONCAT3(x,y,z) CONCAT3p(x,y,z)
#define call_function(func) CONCAT3(func, _, __COUNTER__)()

call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);
call_function(fun);

you need to create an array of function pointers and call through that array.

 typedef int (*pfunc)(int);
 pfunc[] = {func0,func1,func2,func3};
 for(int i = 0; i < 3; i++)
 {
     pfunc[i](arg);
  }

I am sure I have the syntax wrong somewhere - check out http://www.newty.de/fpt/fpt.html

You can't do it like this. Preprocessor is the guy who plays it's role even before program is compiled. What it does is just replaces macros. So when your code gets into compiler there are no macroses at all, they are replaced by actual code.

Take a look at function pointers - that's might help you!

You can't do this. It's simply not part of the C language. What you need to do is create a table that maps the possible inputs to the correct functionality. This can be as simple as a switch statement or you can use a more complicated structure like a hash that uses the input for its keys and has function pointers for values.

Try this (inspired by Bob's comment). You can just index into the fns array instead of using a switch.

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

int fn_0() { printf("0\n"); }
int fn_1() { printf("1\n"); }
int fn_2() { printf("2\n"); }
int fn_3() { printf("3\n"); }
int fn_4() { printf("4\n"); }
int fn_5() { printf("5\n"); }

int main(char **argv, int argc)
{

  int (*fns[6])(void);
  int i;

  fns[0] = *fn_0;
  fns[1] = *fn_1;
  fns[2] = *fn_2;
  fns[3] = *fn_3;
  fns[4] = *fn_4;
  fns[5] = *fn_5;

  for(i = 0; i < 6; ++i)
    fns[i]();

}

Of course if your array is sparse (not a continuous set of integers) then you should probably use a switch.

In response to your issue with function pointers - all you need to do to solve the problem is create a mapping from the ten characters you're interested in to the numbers 1 through 10:

#include <string.h>

typedef int (*funcptr_t)(int);

funcptr_t func_table[11] = { func_unknown, func_ampersand, func_asterisk, func_circumflex, /* ... all the rest here */ };

/* get_index: Returns a value from 1 to 10 if c is in the allowed set, or 0 if not */
static int get_index(char c)
{
    static const char * const lookup_table[] = "&*^><!@#$%";
    char *p = strchr(lookup_table, c);

    return p ? (p - lookup_table) + 1 : 0;
}

Then to call the function, you can just do:

result = func_table[get_index(somechar)](arguments);