c - 为什么我的宏无法编译？

Question

#define LINE_FILE ("Line"#__LINE__"of file"__FILE__)

int main(void)
{
    printf("%s", LINE_FILE);
}

我所期望的：

LINE_FILE = "文件文件名的行号"

并且printf()可以输出这个字符串。

http://ww2.sinaimg.cn/large/005FchG6gw1esb0pvwxr3j30he06jjt7.jpg

Answer 1

__LINE__不会扩展为“字符串”，而是扩展为int.

要解决此问题，您可能需要执行以下操作：

#define _LINE_FILE2(filename, linenumber) "Line " #linenumber " of " filename 
#define _LINE_FILE1(filename, linenumber) _LINE_FILE2(filename, linenumber)
#define LINE_FILE _LINE_FILE1(__FILE__, __LINE__)

---

更多 (gcc) 详细信息

• 预定义宏：https ://gcc.gnu.org/onlinedocs/cpp/Standard-Predefined-Macros.html
• 字符串化：https ://gcc.gnu.org/onlinedocs/cpp/Stringification.html

Answer 2

你需要一个字符串化助手

#define STR_HELPER(x) #x
#define STR(x) STR_HELPER(x)

然后

#define LINE_FILE ("Line" STR(__LINE__)"of file"__FILE__)

Answer 3

AFAIK 你不能在预处理器上这样做，通常的方法是：

    printf("Line %d of file %s", __LINE__, __FILE__);

Answer 4

要调试宏，您可以单独使用 C 预处理器。

行号.c

#define LINE_FILE_WRONG ("Line"#__LINE__"of file"__FILE__)

// solution of LPs
#define STR_HELPER(x) #x
#define STR(x) STR_HELPER(x)
#define LINE_FILE ("Line " STR(__LINE__) " of file " __FILE__)

LINE_FILE_WRONG
LINE_FILE

为这个文件运行 C 预处理器并得到

% cpp linenumber.c
# 1 "linenumber.c"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "/usr/include/stdc-predef.h" 1 3 4
# 1 "<command-line>" 2
# 1 "linenumber.c"







("Line"#8"of file""linenumber.c")
("Line " "9" " of file " "linenumber.c")