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)是否是从 Python 为 32 位或 64 位编译的.我运行的是 64 位 Vista，想确定某个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.

是否有一种仅使用标准 Python 库(目前使用 2.5.4)来执行此操作的简单方法?

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 是 GetBinaryType.您可以使用 pywin32 从 Python 调用它:

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

如果您想在没有 pywin32 的情况下执行此操作，则必须阅读 PE 标头你自己.这是一个例子C#，这里是 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:windowsexplorer.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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