如何从python ctypes中取消引用内存位置？

Question

我想在python ctypes中复制以下c代码：

main() {
  long *ptr = (long *)0x7fff96000000;
  printf("%lx",*ptr);
}

我可以找出如何将此内存位置调用为函数指针，但不只是执行普通的解引用：

from ctypes import *
"""
>>> fptr = CFUNCTYPE(None, None)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.6/ctypes/__init__.py", line 104, in CFUNCTYPE
    class CFunctionType(_CFuncPtr):
TypeError: Error when calling the metaclass bases
    item 1 in _argtypes_ has no from_param method
"""
fptr = CFUNCTYPE(None, c_void_p) #add c_void_p since you have to have an arg
fptr2 = fptr(0x7fff96000000)
fptr2(c_void_p(0))
#python: segfault at 7fff96000000 ip 00007fff96000000

由于存在一个segfault，指令指针指向该内存位置，因此它正在成功调用它。但是我不能让它只读取内存位置：

ptr = POINTER(c_long)
ptr2 = ptr(c_long(0x7fff96000000))
#>>> ptr2[0]
#140735709970432
#>>> hex(ptr2[0])
#'0x7fff96000000'
#>>> ptr2.contents
#c_long(140735709970432)

Answer 1

ctypes.cast。

>>> import ctypes
>>> c_long_p = ctypes.POINTER(ctypes.c_long)
>>> some_long = ctypes.c_long(42)
>>> ctypes.addressof(some_long)
4300833936
>>> ctypes.cast(4300833936, c_long_p)
<__main__.LP_c_long object at 0x1005983b0>
>>> ctypes.cast(4300833936, c_long_p).contents
c_long(42)