在 64 位 python ctypes 中使用 msvcrt
我想使用 ctypes 包从 64 位 python 调用 msvcrt 函数。我显然做错了。正确的做法是否显而易见？
Python 2.7.2 (default, Jun 12 2011, 14:24:46) [MSC v.1500 64 bit (AMD64)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>> import ctypes
>>> libc = ctypes.cdll.msvcrt
>>> fp = libc.fopen('text.txt', 'wb') #Seems to work, creates a file
>>> libc.fclose(ctypes.c_void_p(fp))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
WindowsError: exception: access violation reading 0xFFFFFFFFFF082B28
>>>
如果这段代码满足了我的要求，它将打开和关闭一个文本文件而不会崩溃。
I want to call msvcrt functions from 64-bit python using the ctypes package. I'm obviously doing it wrong. Is the right way to do it obvious?
Python 2.7.2 (default, Jun 12 2011, 14:24:46) [MSC v.1500 64 bit (AMD64)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>> import ctypes
>>> libc = ctypes.cdll.msvcrt
>>> fp = libc.fopen('text.txt', 'wb') #Seems to work, creates a file
>>> libc.fclose(ctypes.c_void_p(fp))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
WindowsError: exception: access violation reading 0xFFFFFFFFFF082B28
>>>
If this code did what I want, it would have opened and closed a text file without crashing.
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默认的ctypes结果类型是32位整数，但文件句柄是指针宽度，即64位。因此，您将丢失文件指针中一半的信息。
在调用 fopen 之前，您必须声明结果类型是指针：
The default ctypes result type is a 32 bit integer but a file handle is pointer width, i.e. 64 bits. You are therefore losing half of the information in the file pointer.
Before you call fopen you must state that the result type is a pointer: