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by jsd1982
1201 days ago
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Very nice description of the bitwise maths, however I'm curious what lead to the need for `(63 - nlz) / 7` in the first place. A divisor of 7 is an odd thing to find when dealing with power-of-2-sized integers. Maybe this expression is used in some sort of expected length calculation to find the number of bytes which would be used for preallocating buffers? In most variable length encoder loops I've seen, you would just do a few bit twiddles to compute your encoded byte values and your exit condition should be a trivial check if there are remaining non-zero bits. |
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`(63 - nlz) / 7` presumably tells you how many bytes you need to read for the varint. The length could be signaled by the number of leading zeros in the first byte (nlz), and each subsequent byte could set a high continuation bit (à la UTF-8 and protobuf varints), thus providing 7 bits of information per byte.
I'm totally speculating but this is generally what I'd expect from the expression `(63 - nlz) / 7` in the context of varints. Continuation bits and leading-zero-counters are redundant but this wouldn't be the first time I've seen something encode redundant information.