> it has to be feeding exactly that many bytes to the SHA3 algorithm

Well, yes, but this is supposed to be a buffer overflow — the algorithm itself is reading and/or writing past the end of the buffer it's been handed. My hypothesis was that Golang is allocating in such a way that reading/writing a single byte past the received slice bounds won't result in a protection fault in the way you'd expect if the allocation were exact.

The buffer overflow is in the C version of the algorithm (and likely related to loop condition checks idiomatic to C's for-loops). The Go version is a fresh implementation, not a wrapping of the C version. I'm no Go programmer, but if I'm not mistaken, the Go implementation just eats little slices of the input buffer until no more buffer is left, leaving all the overflow-danger to the Go array implementation:

https://github.com/golang/crypto/blob/642fcc37f5043eadb2509c...

Possibly not as fast as C, but easier to reason about, I'd say.

Just for what it's worth, we're talking about the amd64 assembly version of the SHA3 code in the x/crypto/sha3 library; I didn't look carefully at how it's called, so if it's used incrementally the way this Go code shows, then yeah, it's fine regardless.

A second later

Oh, wait, yeah, this is just the Keccak permutation in assembly, not the entire hash. That was dumb of me. Yeah, this code looks ok?

Indeed, the two assembly variants are just of the 1600 byte block bit fiddling (keccakF1600).

The outer loop of the Write method (function?) is the same for all architectures.