获取Python中共享库的绝对路径

假设我想在Python中使用libc。这可以很容易地做到

from ctypes import CDLL
from ctypes.util import find_library

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

现在，我知道我可以使用ldconfig获取libc的abspath，但是有没有办法从CDLL对象获取它？它的

\u手柄

是否可以执行某些操作

更新：好的

---

我需要重新定义

链接映射

结构，然后

此上下文中的句柄基本上是对内存映射库文件的引用

然而，在操作系统功能的帮助下，现有的方法可以实现您想要的功能

窗口： Windows为此提供了一个名为

GetModuleFileName

的API。已经有了一些使用示例

linux: 有一个用于此目的的

dlinfo

功能，请参阅

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我使用了ctypes，下面是我针对基于linux的系统的解决方案。到目前为止，我对ctypes一无所知，如果有任何改进建议，我将不胜感激

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)

---

更直接的是：lmptr=指针（LINKMAP）（。。。；abspath=lmptr.contents.l_name；

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)