使用mmap打開文件失敗

Question

Windows上的Python 2.7。試圖使用mmap模塊，但使用打開文件處理程序而不是with open(filename, "r+b") as f:我只是打開它並得到一個WindowsError [Error 5]。

它重現，要麼以管理員身份運行，要麼不以管理員身份運行。 使用codecs.open()不能解決問題。

# -*- coding: utf8 -*- 
from __future__ import print_function 
import mmap 

class QSHFile(object): 
    def __init__(self, filename): 
     self.filename = filename 
     self.file = open(filename, 'r') 
     self.fileno = self.file.fileno() 
     self.mm = mmap.mmap(self.fileno, 0) 
     print(self.mm[:5]) # prints first 5 


if __name__ == '__main__': 
    qsh = QSHFile('example.qsh') 

位或研究後，我到了：

#! /usr/bin/python 
# -*- coding: utf8 -*- 
from __future__ import print_function 
from mmap import ACCESS_READ, mmap 

class QSHFile(object): 
    def __init__(self, filename): 
     self.filename = filename 
     self.file = open(filename, 'rb') 
     self.fileno = self.file.fileno() 
     self.mm = mmap(self.fileno, 0, access=ACCESS_READ) 
     print(self.mm[:5]) 

if __name__ == '__main__': 
    qsh = QSHFile('example.qsh') 

而現在它的正常工作。我現在在做對嗎？

Answer 1

是的，就是這樣，謝謝大家！

# -*- coding: utf8 -*- 
from __future__ import print_function 
from mmap import ACCESS_READ, mmap 
from binascii import hexlify as hex 

class QSHFile(object): 
    def __init__(self, filename): 
     self.filename = filename 
     self.file = open(filename, 'rb') 
     print('File [%s] opened' % self.filename) 
     self.fileno = self.file.fileno() 
     self.mm = mmap(self.fileno, 0, access=ACCESS_READ) 
     print('File size: %s bytes' % self.mm.size()) 
     print(hex(self.mm[:5])) 


if __name__ == '__main__': 
    qsh = QSHFile('example.qsh')