Get absolute path of shared library in Python

Question

Let's say I wanted to use libc in Python. This can be easily done by

from ctypes import CDLL
from ctypes.util import find_library

libc_path = find_library('c')
libc = CDLL(libc_path)

Now, I know I could use ldconfig to get libc's abspath, but is there a way to acquire it from the CDLL object? Is there something that can be done with its _handle?

Update: Ok.

libdl = find_library('dl')
RTLD_DI_LINKMAP = 2
//libdl.dlinfo(libc._handle, RTLD_DI_LINKMAP, ???)

I need to redefine the link_map struct then?!

Answer 1

A handle in this context is basically a reference to the memory mapped library file.

However there are existing ways to achieve what what you want with the help of OS functions.

windows: Windows provides an API for this purpose called GetModuleFileName. Some example of usage is already here.

linux: There is existing a dlinfo function for this purpose, see here.

---

I played around with ctypes and here is my solution for linux based Systems. I have zero knowledge of ctypes so far, if there are any suggestions for improvement I appreciate them.

from ctypes import *
from ctypes.util import find_library

#linkmap structure, we only need the second entry
class LINKMAP(Structure):
    _fields_ = [
        ("l_addr", c_void_p),
        ("l_name", c_char_p)
    ]

libc = CDLL(find_library('c'))
libdl = CDLL(find_library('dl'))

dlinfo = libdl.dlinfo
dlinfo.argtypes  = c_void_p, c_int, c_void_p
dlinfo.restype = c_int

#gets typecasted later, I dont know how to create a ctypes struct pointer instance
lmptr = c_void_p()

#2 equals RTLD_DI_LINKMAP, pass pointer by reference
dlinfo(libc._handle, 2, byref(lmptr))

#typecast to a linkmap pointer and retrieve the name.
abspath = cast(lmptr, POINTER(LINKMAP)).contents.l_name

print(abspath)