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If i==j==k then they’re not disjoint.
by thetallstick 5y ago
If i==j==k then they’re not disjoint.
- morelisp 5y agoThen they're both zero length and a zero-length set is disjoint from all other sets. Anyone want to explain why Rust needs to explicitly de-alias this yet? What a useless digression...
- tialaramex 5y agoWhy LLVM needs this explaining? Probably somewhere an optimiser could inspect something and it doesn't. Rust cuts the slice in half and swaps between the halves, so probably LLVM doesn't convince itself that if you did those swaps directly they aren't ever aliased, but once there are two slices which can't overlap it can see it's fine.