Determine if an executable (or library) is 32 -or 64-bits (on Windows)
If you're running Python 2.5 or later on Windows, you could also use the Windows API without pywin32 by using ctypes.
from ctypes import windll, POINTER
from ctypes.wintypes import LPWSTR, DWORD, BOOL
SCS_32BIT_BINARY = 0 # A 32-bit Windows-based application
SCS_64BIT_BINARY = 6 # A 64-bit Windows-based application
SCS_DOS_BINARY = 1 # An MS-DOS-based application
SCS_OS216_BINARY = 5 # A 16-bit OS/2-based application
SCS_PIF_BINARY = 3 # A PIF file that executes an MS-DOS-based application
SCS_POSIX_BINARY = 4 # A POSIX-based application
SCS_WOW_BINARY = 2 # A 16-bit Windows-based application
_GetBinaryType = windll.kernel32.GetBinaryTypeW
_GetBinaryType.argtypes = (LPWSTR, POINTER(DWORD))
_GetBinaryType.restype = BOOL
def GetBinaryType(filepath):
res = DWORD()
handle_nonzero_success(_GetBinaryType(filepath, res))
return res
Then use GetBinaryType just like you would with win32file.GetBinaryType.
Note, you would have to implement handle_nonzero_success, which basically throws an exception if the return value is 0.
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
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()