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

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

如上的预处理命令，希望
LINE_FILE = "Line 行数 of file 文件名"
然而编译器报错，也不知道错在哪。。。

10 years, 1 month ago

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lgclan 10 years, 1 month ago

Answer 1

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__LINE__ 是int，不支持这种用法。

answered 10 years, 1 month ago

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丶灬11区住民 answered 10 years, 1 month ago

Answer 2

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#define LINE_FILE ("Line # __LINE__ of file __FILE__")
就算可以用也应该是这种格式啊! __LINE__ 和 __FILE__ 这个是define 会直接提换!

补充:
在c艹中:
#define LINE_FILE std::cout<<"Line # " << __LINE__ << " of file "<<__FILE__<<std::endl;
这个宏定义是可以用的!
直接使用 : LINE_FILE

#define LINE_FILE "Line # " << __LINE__ << " of file "<<__FILE__
换成这样的话需要
这样使用 : std::cout<< LINE_FILE <<std::endl;

answered 10 years, 1 month ago

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skgood answered 10 years, 1 month ago