我有以下宏(跟进:When writing a macro in C, how do I find the type and printf specifier for an argument?):
#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 似乎要求解析为char *
,而不是简单地将"%lu"
替换为"required: %s != %s, reality: %s == " mu_format_specifier(actual),
我希望得到"required: %s != %s, reality: %s == " "%lu"
_Generic
C将理解为一个文字字符串。
我可以看到两个可怕的工作:
为每种类型制作mu_assert_equal版本的版本,并使用const
调用正确的版本。
snprintf格式化字符串,但如果格式化字符串必须为_Generic
,我可能会遇到问题。
有更好的选择吗?
(问题是"%lu"
实际上是在评估表达式,而不是简单地替换源字符 - 即"required: %s != %s, reality: %s == "
成为它自己的文字字符串,而不仅仅是生成源字符。与{{1}}统一。)
答案 0 :(得分:0)
我做到了:
#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)
解决方案是使整个格式化字符串成为_Generic
的输出表达式。
从测试代码:
mu_assert_equal(bytes_parsed, 1);
我得到输出:
required: bytes_parsed != 1, reality: bytes_parsed == 0