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valtree's to_le() actually produces different result on big-endian from little-endian. #103183
Description
For commit https://github.com/rust-lang/rust/pull/96591/commits/e14b34c386ad2809e937e0e6e0379c5cc5474954 , the two `to_le()` actually produces different result on big-endian from little-endian.
// Note: Don't use `StableHashResult` impl of `u64` here directly, since that
// would lead to endianness problems.
let hash: u128 = hasher.finish();
let hash_short = (hash.to_le() as u64).to_le();Assume
hash: u128=0x5e0f03940fda80bb6348c650c7b26618, then on LE,hash_short: u64 = 0x6348c650c7b26618, while BEhash_short: u64 = 0x5e0f03940fda80bb, which fails some unit tests on big endian targets because of hash mismatch.
If it's expected behavior to make hash value the same on BE/LE, code here should remove two
to_lecalls. We don't need to adjust endianness unless converting between multi-byte integer and byte sequences.
Originally posted by @ecnelises in #96591 (comment)
Activity
Rageking8 commentedon Oct 18, 2022
@rustbot label +C-bug
Auto merge of rust-lang#103231 - ecnelises:le_fix, r=lcnr
ecnelises commentedon Nov 23, 2022
Can we close this now? (I forgot to mention
Fixed xxxxin commit msg to trigger the bot) Not sure if I have the permission@rustbot close
rustbot commentedon Nov 23, 2022
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Auto merge of rust-lang#103231 - ecnelises:le_fix, r=lcnr