如何获取g ++列出所有包含的文件的路径 [英] How To Get g++ to list paths to all #included files

问题描述

我想让g ++/gcc告诉我通往所有非系统性路径的路径，它是C ++构建中的#include-ing.原来，这是一项艰巨的搜索，因为Google会以十种不同的方式解读它.

I would like to have g++/gcc tell me the paths to everything non-system it is #include-ing in C++ build. Turns out, that is a tough search as Google mus-interprets it about ten different ways.

我想要这些文件名和路径，以便将它们添加到Exuberant CTAGS的搜索路径中.我们有一个庞大的项目，如果我在整个过程中都使用ctags，则生成标记文件大约需要半小时，而编辑器进行查找所需的时间也差不多.

I want these filenames and paths so I can add them to the search path for Exuberant CTAGS. We have a huge project and if I use ctags on the whole thing it takes about half an hour to generate the tags file and nearly as long for the editor to do a look-up.

我们使用CMakeLisats进行编译.如果有指令，我可以将其粘贴到CMakeLists.txt中，那就太妙了.

We use CMakeLisats to do the compiling. If there is a directive I can paste into the CMakeLists.txt, that would be extra wonderfulness.

我真的不需要默认路径和文件名，Johnathan Wakely为此提供了一个很好的工具这里.我认为这几乎涵盖了一个交叉编译工作这一事实.我也不需要跨系统文件.

I don't really need the default paths and filenames, Johnathan Wakely gave a good tool for that here. I think that pretty much covers the fact that this is a cross compile job. I don't need the cross-system files either.

推荐答案

您需要使用-M选项调用g ++.

You need to invoke g++ with the -M option.

手册:

输出规则，而不是输出预处理结果 适用于描述主源文件的依赖关系. 预处理器输出一个包含目标文件名的生成规则 对于该源文件，冒号以及所有包含的名称 文件，包括来自-include或-imacros命令行的文件 选项.

Instead of outputting the result of preprocessing, output a rule suitable for make describing the dependencies of the main source file. The preprocessor outputs one make rule containing the object file name for that source file, a colon, and the names of all the included files, including those coming from -include or -imacros command line options.

值得阅读手册，以考虑其他-M子选项(尤其是-MM和-MF可能有用).

It's worth reading the manual to consider the other -M sub options (-MM and -MF in particular may be of use).

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