循环宏varargs值

我想不出…

但是,如果您的应用程序正在处理同一类型的可变数量的参数,例如:

foo(0);
foo(10, 20, 30);
foo(1, 2, 3, 4, 5, 6, 7, 8, 9);

你不介意用一个函数来帮助你,然后有一些有用的技巧可以做。

这里的主要问题是,不能将这些参数直接传递给varargs函数,因为该函数无法知道要读取多少参数。还有那个问题 可以 用一些预处理器魔法解决:

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

#if defined(__STDC_VERSION__) && (__STDC_VERSION__ >= 199901L)

/* C99-style: anonymous argument referenced by __VA_ARGS__, empty arg not OK */

# define N_ARGS(...) N_ARGS_HELPER1(__VA_ARGS__, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0)
# define N_ARGS_HELPER1(...) N_ARGS_HELPER2(__VA_ARGS__)
# define N_ARGS_HELPER2(x1, x2, x3, x4, x5, x6, x7, x8, x9, n, ...) n

# define foo(...) foo_helper(N_ARGS(__VA_ARGS__), __VA_ARGS__)

#elif defined(__GNUC__)

/* GCC-style: named argument, empty arg is OK */

# define N_ARGS(args...) N_ARGS_HELPER1(args, 9, 8, 7, 6, 5, 4, 3, 2, 1)
# define N_ARGS_HELPER1(args...) N_ARGS_HELPER2(args)
# define N_ARGS_HELPER2(x1, x2, x3, x4, x5, x6, x7, x8, x9, n, x...) n

# define foo(args...) foo_helper(N_ARGS(args), args)

#else

#error variadic macros for your compiler here

#endif

static inline void foo_helper(unsigned int n_args, ...)
{
  unsigned int i, arg;
  va_list ap;

  va_start(ap, n_args);
  printf("%u argument(s):\n", n_args);
  for (i = 0; i < n_args; i++) {
    arg = va_arg(ap, unsigned int);
    printf("  %u\n", arg);
  }
  va_end(ap);
}

int main(void)
{
  foo(0);
  foo(10, 20, 30);
  foo(1, 2, 3, 4, 5, 6, 7, 8, 9);
  return 0;
}

输出:

$ gcc -W -Wall -std=c99 -pedantic -o va_macro va_macro.c
$ ./va_macro
1 argument(s):
  0
3 argument(s):
  10
  20
  30
9 argument(s):
  1
  2
  3
  4
  5
  6
  7
  8
  9
$