Question: why consider only 7 bits per byte for an unsigned int? about go-varint HOT 3 CLOSED

A varint is somewhat like utf8: all "wire-bytes" but the last have the highest bit set, to indicate "more is coming". Hence the "wiresize" of a varint representation is the "payload" divided into chunks of 7 bits.

For reference this is what an actual varint encoder looks like: https://golang.org/src/encoding/binary/varint.go?s=1611:1652#L31

func PutUvarint(buf []byte, x uint64) int {
	i := 0
	for x >= 0x80 {
		buf[i] = byte(x) | 0x80
		x >>= 7
		i++
	}
	buf[i] = byte(x)
	return i + 1
}

from go-varint.

Got it. Thanks for clearing that up.

from go-varint.

@pfac just a note that this kind of varint encoding also goes by the name LEB128, so there might already be an implementation available for your language of choice.

from go-varint.