The book Advanced Linux Programming contains:
The system searches only /lib and /usr/lib, by default. If a shared library that is linked into your program is installed outside those directories, it will not be found, and the system will refuse to run the program.
One solution to this problem is to use the -Wl, -rpath option when linking the program. Another solution to this problem is to set the LD_LIBRARY_PATH environment variable when running the program.
So from information above, I assume adding path to $LD_LIBRARY_PATH is equivalent to passing argument to linker when compiling something like -Wl,-rpath,SOME_PATH until I hit this problem:
I am trying to generate a shared library and this shared library depends on some .so files somewhere at my system. Here's the command for compilation:
g++ -fPIC -shared -Wl,-soname,libA.so -o libA.so SRC.cpp -L PATH_A -I PATH_B -Wl,-rpath,PATH_C:PATH_D:PATH_E -la -lb -lc
After I compile, I use ldd to check if there's any dependency issue and something unexpected to me happens:
Some of dependency are found and some of them are not (libA.so => not found), however all the paths it needs (PATH_C PATH_DPATH_E) have been passed to argument -Wl,-rpath.
So it seems -Wl,-rpath does not work for this case , the only solution to make it work is to add the path to libA.so to $LD_LIBRARY_PATH and ldd returns that all the dependency are found.
I would like to know why -Wl,-rpath fails in, what's the difference between adding the path to $LD_LIBRARY_PATH and passing an argument to -Wl,-rpath?