从Python中的二进制文件中读取整数 [英] Reading integers from binary file in Python

问题描述

我想读 BMP在Python 文件。我知道前两个字节
表明BMP公司。在接下来的4个字节是文件大小。当我excecute：

I'm trying to read a BMP file in Python. I know the first two bytes indicate the BMP firm. The next 4 bytes are the file size. When I excecute:

fin = open("hi.bmp", "rb")
firm = fin.read(2)  
file_size = int(fin.read(4))  

我得到

ValueError错误：无效的字面INT（）基数为10：'F＃\\ X13

ValueError: invalid literal for int() with base 10: 'F#\x13'

我想要做的就是读那些四个字节作为一个整数...看来Python的是阅读他们的字符并返回一个字符串，它不能转换为整数。我怎么能这样做是否正确？

What I want to do is reading those four bytes as an integer... It seems Python is reading them as characters and returning a string, which cannot be converted to an integer. How can I do this correctly?

推荐答案

的读方法返回字节字符串序列。从字符串的字节序列为二进制数据转换，使用内置的结构模块：的 http://docs.python.org/library/struct.html 。

The read method returns a sequence of bytes as a string. To convert from a string byte-sequence to binary data, use the built-in struct module: http://docs.python.org/library/struct.html.

import struct

print(struct.unpack('i', fin.read(4)))

注意解压总是返回一个元组，因此 struct.unpack（'我'，fin.read（4））[0] 给你是后的整数值。

Note that unpack always returns a tuple, so struct.unpack('i', fin.read(4))[0] gives the integer value that you are after.

您应该使用格式字符串'＆LT;我（小于是一个修饰符，表示little-endian字节顺序和标准大小和对齐方式 - 默认是使用平台的字节顺序，大小和取向）。按照BMP格式规范，字节应该写在英特尔/ little-endian字节顺序。

You should probably use the format string '<i' (< is a modifier that indicates little-endian byte-order and standard size and alignment - the default is to use the platform's byte ordering, size and alignment). According to the BMP format spec, the bytes should be written in Intel/little-endian byte order.

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