确定可执行文件（或库）是32位还是64位（在Windows上） [英] Determine if an executable (or library) is 32 -or 64-bits (on Windows)

问题描述

我试图找出给定的可执行文件（或库）是从Python编译为32位还是64位。我正在运行64位Vista，并希望确定目录中的某个应用程序是针对32位还是64位编译的。

I am trying to find out if a given executable (or library) is compiled for 32-bits or 64-bits from Python. I am running Vista 64-bits and would like to determine if a certain application in a directory is compiled for 32-bits or 64-bits.

有没有一种简单的方法要仅使用标准Python库（当前使用2.5.4）执行此操作？

Is there a simple way to do this using only the standard Python libraries (currently using 2.5.4)?

推荐答案

Windows API为此 GetBinaryType 。您可以使用 pywin32 从Python调用此函数：

The Windows API for this is GetBinaryType. You can call this from Python using pywin32:

import win32file
type=GetBinaryType("myfile.exe")
if type==win32file.SCS_32BIT_BINARY:
    print "32 bit"
# And so on

如果要在没有pywin32的情况下执行此操作自己阅读 PE标头。这是示例 C＃，这是Python的快速移植：

If you want to do this without pywin32, you'll have to read the PE header yourself. Here's an example in C#, and here's a quick port to Python:

import struct

IMAGE_FILE_MACHINE_I386=332
IMAGE_FILE_MACHINE_IA64=512
IMAGE_FILE_MACHINE_AMD64=34404

f=open("c:\windows\explorer.exe", "rb")

s=f.read(2)
if s!="MZ":
    print "Not an EXE file"
else:
    f.seek(60)
    s=f.read(4)
    header_offset=struct.unpack("<L", s)[0]
    f.seek(header_offset+4)
    s=f.read(2)
    machine=struct.unpack("<H", s)[0]

    if machine==IMAGE_FILE_MACHINE_I386:
        print "IA-32 (32-bit x86)"
    elif machine==IMAGE_FILE_MACHINE_IA64:
        print "IA-64 (Itanium)"
    elif machine==IMAGE_FILE_MACHINE_AMD64:
        print "AMD64 (64-bit x86)"
    else:
        print "Unknown architecture"

f.close()

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