c/c++语言开发共享如何使宏返回字符而不是字符串？

我有以下宏（跟进： 在C中编写宏时，如何找到参数的类型和printf说明符？ ）：

#define mu_format_specifier(expression) _Generic((expression), unsigned long: "%lu", int: "%i") #define mu_assert_equal(actual, expected) do {  if (actual != expected) {  char *message = malloc(MAX_ERROR_MESSAGE_LENGTH);  if (message == NULL) { printf("malloc failed"); exit(1); }  snprintf(message, MAX_ERROR_MESSAGE_LENGTH,  "required: %s != %s, reality: %s == " mu_format_specifier(actual),  #actual, #expected, #actual, actual);  return message;  }  } while (0) 

问题是mu_format_specifier 似乎解析为char *而不是简单地将"%lu"替换为"required: %s != %s, reality: %s == " mu_format_specifier(actual),我想要产生"required: %s != %s, reality: %s == " "%lu" C将理解为一个文字字符串。

我可以看到两个可怕的工作：

还有更好的选择吗？

（问题是_Generic实际上是在计算表达式，而不是简单地替换源字符 – 即"%lu"成为它自己的文字字符串，而不仅仅是生成与"required: %s != %s, reality: %s == "统一的源字符"required: %s != %s, reality: %s == " 。）

 #define mu_assert_equal(actual, expected) do {  if (actual != expected) {  char *message = malloc(MAX_ERROR_MESSAGE_LENGTH);  if (message == NULL) { printf("malloc failed"); exit(1); }  snprintf(message, MAX_ERROR_MESSAGE_LENGTH, _Generic(  (actual),  unsigned long: "required: %s != %s, reality: %s == %lu",  int: "required: %s != %s, reality: %s == %i"  ),  #actual, #expected, #actual, actual);  return message;  }  } while (0) 

 mu_assert_equal(bytes_parsed, 1); 

 required: bytes_parsed != 1, reality: bytes_parsed == 0 

以上就是c/c++开发分享如何使宏返回字符而不是字符串？相关内容,想了解更多C/C++开发(异常处理)及C/C++游戏开发关注计算机技术网(www.ctvol.com)!)。