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Category:Rescuing dogs
Category:Dog breeds originating in Ukraine
Category:Rare dog breedsQ:
Why is the last bit set in each byte in big endian ordering?
Let's say I have a binary file that looks like this:
1111 1110 1101 1000
Each byte represents a unsigned integer.
I would assume, and it makes sense, that the most significant bit is the first bit (0) in the byte.
When I convert the file to little endian and use python to unpack it, I get something like this:
>>> import struct
>>> from struct import unpack
>>> struct.unpack('>i', open('big_endian.bin', 'rb'))
[-115]
Why does it end up with the last bit set?
I know this might be a trivial question, but I have no idea how to google this properly. I can't seem to find any questions on this subject on stackoverflow.
A:
The last bit of each byte is used to fill the end of the buffer. If the end of the buffer is not a multiple of 8 bits, then the bit is ignored. In this case, the last bit of each byte is a value of -32.
If the end of the buffer is a multiple of 8 bits, then it's filled with 1s. In this case, the last bit is set to -1.
Hence the result is -115, which is the same result as you found out.
In the end, it's just a convention and it's not even specified anywhere.
A:
This is a convention, not a standard.
Endianness is a property of how bytes are encoded in a file. A byte is a collection of 8 bits. In your example, a byte is a set of 2 bits. A little endian system contains the least significant bit in the first byte (10 in your example), and the most significant bit in the last byte (11 in your example). Big endian systems have the most significant bit in the first byte, and the least significant bit in the last byte.
Parsing the data in that way is convenient for performing operations on the data, but you'll need to make sure that you read the data be359ba680
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