How do I call a function with a variable number of parameters?

How do I call execlp() with a variable number of arguments for different proc开发者_StackOverflow中文版esses?

If you don't know how many arguments you'll need at the time you are writing your code, you want to use execvp(), not execlp():

char **args = malloc((argcount + 1) * sizeof(char *));
args[0] = prog_name;
args[1] = arg1;
...
args[argcount] = NULL;

execvp(args[0], args);

This answers only the title question

From Wikipedia Covers old and new styles

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

void printargs(int arg1, ...) /* print all int type args, finishing with -1 */
{
  va_list ap;
  int i;

  va_start(ap, arg1); 
  for (i = arg1; i != -1; i = va_arg(ap, int))
    printf("%d ", i);
  va_end(ap);
  putchar('\n');
}

int main(void)
{
   printargs(5, 2, 14, 84, 97, 15, 24, 48, -1);
   printargs(84, 51, -1);
   printargs(-1);
   printargs(1, -1);
   return 0;
}

execlp() can be called with variable number or arguments, so just call:

int ret;
ret = execlp("ls", "ls", "-l", (char *)0);
ret = execlp("echo", "echo", "hello", "world", (char *)0);
ret = execlp("man", "man", "execlp", (char *)0);
ret = execlp("grep", "grep", "-l", "pattern", "file1", "file2", (char *)0);

Execlp already as a variable number of parameters. What do you want to do exactly ? You can probably a variadic macro :

#define myfind(...) execlp("find", "find", __VA_ARGS__)

This is a rather useless example, but without knowing more precisely what you wanted to do, that's all I could come up with