How can I find the full file path given a library name like libfoo.so.1?

Question

Without implementing a linker or using ldd, how can I find the full path to a library? Is there a standard library available for that on Linux? (POSIX maybe?)

Using ldd and grep on a file that is knowingly using libGL.so.1, it looks like:

$ ldd /usr/bin/glxinfo | grep libGL
libGL.so.1 => /usr/lib/libGL.so.1 (0x00007f34ff796000)

Given a library name like libGL.so.1, how can I find the full path /usr/lib/libGL.so.1?. Preferably accepting an option for finding 32-bit and 64-bit libraries. If no library does that, does a program exist to do this? Something like find-library-path libGL.so.1. The locate libGL.so.1 command does not count.

I don't want to actually load the library using dlopen or something if it executes code from that library.

Solution 1

Use ldconfig which is the tool that manages link space.

The -p flag lets you browse all available linkable libraries.

Solution 2

Expanding on Honky Tonk's answer, the command echo "$(ldconfig -p | grep libGL.so.1 | tr ' ' '\n' | grep /)" will give you the path alone.

Solution 3

If you don't mind actually loading the library and using some nonstandard but widely-available functions, calling dladdr on any symbol from the library will return information containing the full pathname that was loaded.

Solution 4

For systems with GNU libc and Python the following is the closest I found. It uses LD_DEBUG (described in the man page of ld.so(8)).

LD_DEBUG=libs python3 -c "import ctypes; ctypes.CDLL('libssl.so.1.0.0')" 2>&1 | \
    grep -A 1000 "initialize program: python" | grep -A 3 "find library"

The output (for libssl.so.1.0.0) is the following:

 15370: find library=libssl.so.1.0.0 [0]; searching
 15370:  search cache=/etc/ld.so.cache
 15370:   trying file=/lib/x86_64-linux-gnu/libssl.so.1.0.0
 15370: 
 15370: find library=libcrypto.so.1.0.0 [0]; searching
 15370:  search cache=/etc/ld.so.cache
 15370:   trying file=/lib/x86_64-linux-gnu/libcrypto.so.1.0.0
 15370: 

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