long 的结构长度取决于字节顺序？

Question

I want to use struct.unpack() to get a long value from a byte string in Python 2.7, but I found some strange behaviour and I would like to know if this is a bug or not, and also, what I can do to get around it.

What happens is the following:

import struct
>>> struct.unpack("L","")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
struct.error: unpack requires a string argument of length 8

This is the expected behaviour. It wants an 8-byte string to extract the 8-byte long value.

Now I change the byte order to network byte order and I get the following:

>>> struct.unpack("!L","")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
struct.error: unpack requires a string argument of length 4

Suddenly it only wants 4 bytes for the 8-byte long value.

What is up with that, and what can I do to get around this?

Answer 1

The issue here is that struct is using your machine's native long size in the first case, and struct 's "standard" long size in the second.

In the first example, you don't specify a byte-order, size, and alignment specifier (one of @=<>! ) so struct assumes '@' , which uses your machine's native values, namely your machine's native long size. In order to be consistent across platforms,struct defines standard sizes for each type , which may differ from your machine's native sizes. Every specifier except for '@' uses these standard sizes.

So, in the case of long , struct 's standard is 4 bytes, which is why '!L' (or any '[=<>!]L' expects 4 bytes. However, your machine's native long is apparently 8 bytes, which is why '@L' or 'L' expects 8. If you want to use your machine's native byte-order, but still be compatible with struct 's standard sizes, I would recommend that you specify all formats with '=' , instead of letting Python default it to '@' .

---

You can check the size expectation with struct.calcsize , ie:

>>> struct.calcsize('@L'), struct.calcsize('=L')

(this returns (4, 4) on my 64-bit Windows 10 machine, but (8, 4) on my 64-bit Ubuntu 16.04 machine.)

You can also directly check your machine's type size by compiling a C script to check the value of sizeof(long) , for example:

#include <stdio.h>

int main()
{
    printf("%li",sizeof(long));
    return 0;
}

I followed this guide to compile the C script on my Windows 10 machine.