C++ g++ 预处理器输出
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g++ preprocessor output
提问by mirt
I want to get only preprocessed version of file.cc. I did g++ -E file.cc, got:
我只想获得file.cc. 我做了g++ -E file.cc，得到：
# 1 "file.cc"
# 1 "<command-line>"
# 1 "file.cc"
What did I do wrong?
我做错了什么？
回答by Andreas Fester
Assumed that your source file contains a simple main function:
假设您的源文件包含一个简单的 main 函数：
$ cat file.cc
int main() {
return 0;
}
Then, with the command you showed, the output looks like this:
然后，使用您显示的命令，输出如下所示：
$ g++ -E file.cc
# 1 "file.cc"
# 1 "<built-in>"
# 1 "<command-line>"
# 1 "file.cc"
int main() {
return 0;
}
The output which is shown in the question happens when file.ccis empty. Note that "empty" means that the file can still contain comments, or #ifdefblocks with a condition which evaluates to false - since the preprocessor filters them out, they do not appear in the output either.
问题中显示的输出在file.cc为空时发生。请注意，“空”意味着文件仍然可以包含注释或#ifdef条件为 false 的块 - 由于预处理器将它们过滤掉，它们也不会出现在输出中。
回答by Christian Stieber
If you don't want the line markers, use the -P option:
如果您不想要线条标记，请使用 -P 选项：
-P
Inhibit generation of linemarkers in the output from the preprocessor. This might
be useful when running the preprocessor on something that is not C code, and will
be sent to a program which might be confused by the linemarkers.
