python中的解包函数
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unpack function in python
提问by Neon Flash
I am trying to understand the unpack function in Python and how it uses the format string.
我试图了解 Python 中的解包函数以及它如何使用格式字符串。
I am using the format string, "I" which corresponds to unsigned int (size, 4 bytes) as an example.
我使用格式字符串“I”作为示例，它对应于 unsigned int（大小，4 个字节）。
As per the documentation, the unpack function will accept a string and convert it to a list of values based on the format string.
根据文档，解包函数将接受一个字符串并将其转换为基于格式字符串的值列表。
http://docs.python.org/2/library/struct.html
http://docs.python.org/2/library/struct.html
So, I used the input value as a string, "test" and here is the output:
因此，我将输入值用作字符串“test”，这是输出：
>>> import struct
>>> input="test"
>>> l = struct.unpack("I", input)[0]
>>> print l
1953719668
I am trying to understand how the output value was derived from the input.
我试图了解输出值是如何从输入中导出的。
>>> from struct import *
>>> calcsize('I')
4
size of 'I' is 4 bytes. the string, "test" has 4 characters which is 4 bytes. I tried converting each character to its corresponding Hex ASCII value and storing it in little endian order but it does not match the output above.
'I' 的大小是 4 个字节。字符串“test”有 4 个字符，即 4 个字节。我尝试将每个字符转换为其相应的十六进制 ASCII 值并以小端顺序存储它，但它与上面的输出不匹配。
Any help would be appreciated.
任何帮助，将不胜感激。
采纳答案by falsetru
Use 4sif you want unpack string as is.
4s如果您想按原样解压缩字符串，请使用。
>>> struct.unpack('4s', 'test')[0]
'test'
1953719668 is derived by: (little endian)
1953719668 衍生自：（小端）
>>> ord('t') + (ord('e') << 8) + (ord('s') << 16) + (ord('t') << 24)
1953719668
