如何知道 libpython27.a 是 32 位還是 64 位？ (How to know whether libpython27.a is 32-bit or 64-bit?)

問題描述

如何知道 libpython27.a 是 32 位還是 64 位？ (How to know whether libpython27.a is 32‑bit or 64‑bit?)

我有一個 libpython27.a 文件：在 Windows 7 x64 上如何知道它是 32 位還是 64 位？

參考解法

方法 1:

Try dumpbin /headers "libpython27.a". (dumpbin reference)

The output will contain

FILE HEADER VALUES 14C machine (x86)

or

FILE HEADER VALUES 8664 machine (x64)

Note that if you get an error message like:

E:\temp>dumpbin /headers "libpython27.a"
LINK: extra operand `libpython27.a'
Try `LINK ‑‑help' for more information.

It means there is a copy of the GNU link utility somewhere in the search path. Make sure you use the correct link.exe (e.g. the one provided in C:\Program Files (x86)\Common Files\Microsoft\Visual C++ for Python\9.0\VC\bin). It also requires mspdb80.dll, which is in the same folder or something in PATH, otherwise you'll get the error message:

方法 2:

When starting the Python interpreter in the terminal/command line you may also see a line like:

Python 2.7.2 (default, Jun 12 2011, 14:24:46) [MSC v.1500 64 bit (AMD64)] on win32

Where [MSC v.1500 64 bit (AMD64)] means 64‑bit Python.

Or

Try using ctypes to get the size of a void pointer:

import ctypes
print ctypes.sizeof(ctypes.c_voidp)

It'll be 4 for 32 bit or 8 for 64 bit.

方法 3:

On Linux you can use: objdump ‑a libpython27.a|grep 'file format'.

Example:

f@f‑VirtualBox:/media/code$ objdump ‑a libpython27.a|grep 'file format'
dywkt.o:     file format pe‑i386
dywkh.o:     file format pe‑i386
dywks01051.o:     file format pe‑i386
dywks01050.o:     file format pe‑i386
dywks01049.o:     file format pe‑i386
dywks01048.o:     file format pe‑i386
[...]

(by Franck Dernoncourt、cgohlke、Kumar Saurabh、Franck Dernoncourt)

參考文件

1. How to know whether libpython27.a is 32‑bit or 64‑bit? (CC BY‑SA 2.5/3.0/4.0)