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They're doing 32-bit reversal on a byte?.....
Yes, and it makes sense. Imo a comment that explains why this is so would be much more useful.
So the variable byte isn't a uint8_t, but a uint32_t?
Well, "byte" is an unsigned int and the code assumes that this is at least 32 bits wide. It contains a message byte in its 8 LSB bits, after that line the reflection of that byte is in the 8 MSB bits of "byte".
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