在python中使用结构包
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Using struct pack in python
提问by user2578666
I have a number in integer form which I need to convert into 4 bytes and store it in a list . I am trying to use the struct module in python but am unable to get it to work:
我有一个整数形式的数字，我需要将其转换为 4 个字节并将其存储在列表中。我正在尝试在 python 中使用 struct 模块，但无法让它工作：
struct.pack("i",34);
This returns 0 when I am expecting the binary equivalent to be printed. Expected Output:
当我期望打印等效的二进制文件时，这将返回 0。预期输出：
[0x00 0x00 0x00 0x22]
But struct.pack is returning empty. What am I doing wrong?
但是 struct.pack 返回空。我究竟做错了什么？
采纳答案by Martijn Pieters
The output is returned as a byte string, and Python will print such strings as ASCII characters whenever possible:
输出作为字节字符串返回，Python 会尽可能将此类字符串打印为 ASCII 字符：
>>> import struct
>>> struct.pack("i",34)
'"\x00\x00\x00'
Note the quote at the start, that's ASCII codepoint 34:
请注意开头的引号，即 ASCII 代码点 34：
>>> ord('"')
34
>>> hex(ord('"'))
'0x22'
If you expected the ordering to be reversed, then you may need to indicate a byte order:
如果您希望顺序颠倒，那么您可能需要指明字节顺序：
>>> struct.pack(">i",34)
'\x00\x00\x00"'
where >indicates big-endian alignment.
其中>表示大端对齐。
