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I have 4 variable between 0~0x1FF(9-bit) and 1 variable between 0~0xF(4-bit) and I want to pack them into binary. As usual, I will pack the first four with pack('H', var)(unsigned short, 16-bit), then pack the last one with pack('B', var)(unsigned char, 8-bit).

so I will cost 72 bits:

16 * 4 + 8 = 72

I am facing a special situation: my storage is very very precious, I just need 40 bits:

4 * 9 + 1 * 4

I waste 32 bits, and also these 32 bits can be important to me, because I have a lot of data to pack, every time I waste 32 bits, at last, my storage complains to me.

In fact, I can change 4 9-bit variable and 1 4-bit variable into 5 8-bit variable(unsigned char), then I pack them all with pack('B', var), of course I save 32 bits.

4 * 9 + 1 * 4 == 5 * 8

How can I simply change pack 4 9-bit variable and 1 4-bit variable int 5 8-bit variable?

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  • 1) Don't spam tags! This is not related to C. 2) Use bitops. Commented Mar 19, 2016 at 17:00

1 Answer 1

Reset to default
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Pack them yourself:

import struct
ninebits = [0b110011001,0b000000000,0b111111111,0b010101010]
fourbit = 0b1010
pack = 0
for i,v in enumerate(ninebits):
    pack <<= 9
    pack |= v
pack <<= 4
pack |= fourbit
print('{:040b}'.format(pack))
print('{:010X}'.format(pack))
print(struct.pack('>LB',pack >> 8, pack & 0xFF))

Output:

1100110010000000001111111110101010101010
CC803FEAAA
b'\xcc\x80?\xea\xaa'

Note that ? is ASCII for \x3f.

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1 Comment

I don't think that it is doable directly with pack without going through the shift and adding process you put in the for loop?...I'm just trying to figure out a way for that...:P...but nevertheless, great answer +1

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