从共享库中删除嵌入的源文件名 [英] Remove embedded source filenames from shared libraries

问题描述

当我从源文件secret.cc构建共享库mylib.so时，生成的共享对象包含源文件名：

do_global_ctors_aux^@secret.cc ^ @ __ DTOR_END ...

但我不想泄露该文件的名称secret.cc）给我的图书馆的用户。有没有办法从共享对象中剥离文件名信息，或者防止它被插入到第一个位置？

解决方案

很简单：不要让编译器从一开始就知道源文件名。取代

  g ++ -std = c ++ 11 -O3 -Wall -c my_source.cc -o my_source.o 
  

执行此操作：

< pre class =lang-none prettyprint-override> cat my_source.cc | g ++ -std = c ++ 11 -O3 -Wall -c -x c ++ - -o my_source.o


请注意，你需要明确提供 -x c ++ ，错误消息显然不会包含文件名，还有一个额外的警告：当你的源包含相对包含，即，包括在引号（ #includefoo.hpp）中，而不是尖括号（ #include< foo.hpp> ），这些将不再工作，因为编译器不能引用该文件的目录，它只是看到一个字节流从管道。

When I build a shared library "mylib.so" from a source file "secret.cc", the resulting shared object contains the source filename:

... do_global_ctors_aux^@secret.cc^@__DTOR_END ...

But I don't want to divulge the name of that file ("secret.cc") to the users of my library. Is there a way to strip the filename information from the shared object, or to prevent it from being inserted in the first place?

解决方案

It's quite simple: Don't let the compiler know the source's filename from the very beginning. Instead of

g++ -std=c++11 -O3 -Wall -c my_source.cc -o my_source.o

do this:

cat my_source.cc | g++ -std=c++11 -O3 -Wall -c -x c++ - -o my_source.o

Note that you need to provide -x c++ explicitly, the error messages obviously won't contain the filename anymore and there is one additional caveat: When your sources contain relative includes, i.e., includes in quotes (#include "foo.hpp") instead of angle brackets (#include <foo.hpp>), those will no longer work as the compiler can't refer to the file's directory, it just sees a byte-stream from a pipe.

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