Shared Library Path as Executable Directory

Question

Hi,

I have an application that is broken into several libraries for purposes of code reuse. On Windows all I have to do is put the .dll files in the same path as the executable and it automatically finds them. On Linux (since it hardcodes the paths to things) I have to specify the environmental variable LD_LIBRARY_PATH or preload the libraries before the executable.

I've seen some things about embedding the path using the linker option of -Wl,-rpath=<PATH> and I've tried it using . as the path. But that just looks in the current working directory, not the executable's directory.

Is there a way to specify in the linker to look in the directory of the executable for the shared libraries by default (like on Windows)?

Thanks! Matt

Answer 1

You need $ORIGIN. See this and this.

Remember that the $ has to really end up in the path, so you need to quote or escape it in the link command line.

Update: You can see what the linker actually put into your executable with

readelf -d /path/to/exe | grep RPATH

Here is what the output should look like:

 0x0000000f (RPATH)              Library rpath: [$ORIGIN]

Answer 2

Wrap your program in a shell script:

#!/bin/sh

PROGRAM_DIRECTORY="`dirname "$0"`"
export LD_LIBRARY_PATH="$PROGRAM_DIRECTORY"

"$PROGRAM_DIRECTORY/program_executable" "$@"

If you run this script (instead of your executable) your program will link just fine.