对于不遵循 libNAME.so 命名约定的库，如何在没有 -l 或硬编码路径的情况下使用 GCC 链接?
我有一个共享库，我希望使用 GCC 链接可执行文件.共享库的名称不是 libNAME.so 形式的非标准名称，因此我不能使用通常的 -l 选项.(它恰好也是一个 Python 扩展，因此没有lib"前缀.)
I have a shared library that I wish to link an executable against using GCC. The shared library has a nonstandard name not of the form libNAME.so, so I can not use the usual -l option. (It happens to also be a Python extension, and so has no 'lib' prefix.)
我可以将库文件的路径直接传递到链接命令行，但这会导致库路径被硬编码到可执行文件中.
I am able to pass the path to the library file directly to the link command line, but this causes the library path to be hardcoded into the executable.
例如:
g++ -o build/bin/myapp build/bin/_mylib.so
有没有办法链接到这个库而不会导致路径被硬编码到可执行文件中?
Is there a way to link to this library without causing the path to be hardcoded into the executable?
推荐答案
有一个:"前缀可以让你给你的库起不同的名字.如果你使用
There is the ":" prefix that allows you to give different names to your libraries. If you use
g++ -o build/bin/myapp -l:_mylib.so other_source_files
应该在路径中搜索 _mylib.so.
should search your path for the _mylib.so.
相关文章