Python struct.Struct.size returns unexpected value

Question

I use Python to convert some files to binary, but I came across an odd network.

Problem

the code

import struct s = struct.Struct('Bffffff') print s.size 

Result

 28 

Obviously, the expected size will be 25 , but it seems to interpret the first byte ( B ) as a 4-byte integer. It will also write out a 4 byte integer instead of a byte.

Work around

A workaround exists, namely splitting B out into a separate struct , for example:

the code

 import struct s1 = struct.Struct('B') s2 = struct.Struct('ffffff') print s1.size + s2.size 

Result

 25 

Are there any explanations for this behavior?

Answer 1

If you do not specify any character for byte order, alignment, struct use your own byte order, alignment ( @ ); which cause filling.

By explicitly specifying the byte order, you can get what you want:

 >>> struct.Struct('!Bffffff').size # network byte order 25 >>> struct.Struct('=Bffffff').size # native byte order, no alignment. 25 >>> struct.Struct('>Bffffff').size # big endian 25 >>> struct.Struct('<Bffffff').size # little endian 25 >>> struct.Struct('@Bffffff').size # native byte order, alignment. (+ native size) 28 

Answer 2

From docs

Filling is only automatically added between successive members of the structure. No additive is added at the beginning or end of the encoded structure.

If you test

 >>> import struct >>> s1 = struct.Struct('B') >>> print s1.size 1 >>> s1 = struct.Struct('f') >>> print s1.size 4 

So, when you add it, 25 ... But on the contrary, B is 1, and the rest 4 , so it will be complemented to make it 4 , so the answer is 28 Consider this example

 >>> s1 = struct.Struct('Bf') >>> print s1.size 8 

Here again, B is 1 , and padded 3 and f are 4 , so it reaches 8 , which is as expected.

As mentioned here , to override it, you will have to use non-native methods

 >>> s1 = struct.Struct('!Bf') >>> print s1.size 5 

No additive is added when using step size and alignment, for example. with '<,'>, '=, and' !.