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- __FILE__ macro manipulation handling at compile time 8 answers
- 编译时__FILE__宏操作处理8个答案
I want to print the filename that is being used in my project. But if I use __FILE__, it prints entire path along with the file name which is long in my case and disturbs the log indents.
我想打印我的项目中使用的文件名。但是如果我使用__FILE__，它会打印整个路径以及文件名，这在我的情况下很长，并且会干扰日志缩进。
Can anyone please help how I can print just the file name using a common parameter across my project.
任何人都可以帮助我如何在我的项目中使用通用参数打印文件名。
2 个解决方案
#1
2
In your C code, you might use instead of __FILE__ the expression basename(__FILE__); however, this has the disadvantage of calling basename at every occurrence.
在您的C代码中，您可以使用表达式basename（__ FILE__）来代替__FILE__;但是，这样做的缺点是每次都会调用basename。
Alternatively, compile path/foo.c with a command like
或者，使用类似命令编译path / foo.c
gcc -Wall -c -g -DBASE_FILE=\"foo\" path/foo.c
(you could have a generic make rule giving that)
（你可以有一个通用的制作规则给出）
then use BASE_FILE instead of __FILE__ in your code.
然后在代码中使用BASE_FILE而不是__FILE__。
With GCC, you could use __BASE_FILE__ instead of __FILE__
使用GCC，您可以使用__BASE_FILE__而不是__FILE__
#2
-1
Consider using basename. I don't think the preprocessor gives you much more than __FILE__.
考虑使用basename。我认为预处理器不会给你提供比__FILE__更多的东西。
#1
2
In your C code, you might use instead of __FILE__ the expression basename(__FILE__); however, this has the disadvantage of calling basename at every occurrence.
在您的C代码中，您可以使用表达式basename（__ FILE__）来代替__FILE__;但是，这样做的缺点是每次都会调用basename。
Alternatively, compile path/foo.c with a command like
或者，使用类似命令编译path / foo.c
gcc -Wall -c -g -DBASE_FILE=\"foo\" path/foo.c
(you could have a generic make rule giving that)
（你可以有一个通用的制作规则给出）
then use BASE_FILE instead of __FILE__ in your code.
然后在代码中使用BASE_FILE而不是__FILE__。
With GCC, you could use __BASE_FILE__ instead of __FILE__
使用GCC，您可以使用__BASE_FILE__而不是__FILE__
#2
-1
Consider using basename. I don't think the preprocessor gives you much more than __FILE__.
考虑使用basename。我认为预处理器不会给你提供比__FILE__更多的东西。