Eq for MemoBytes should compare the bytes. #2565
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Two instances of a memoised type are equal iff they have the same bytes. Justification: - We require that deserialisation is left inverse to serialisation. So if we have `serialise x == serialise y` `deserialise (serialise x) == deserialise (serialise y)` `x == y` - `Eq` should be compatible with hashing (e.g. if `a == b`, then `hash a == hash b`). Currently, this is broken. If I serialise a `TxBodyRaw` in two different valid ways (as is perfectly allowed), then the resulting `TxBody` will currently compare as equal but hash to a different value. - In choosing not to require a canonical serialisation, the serialised bytes are an integral part of the type, and must be considered in comparing for equality.
KtorZ
reviewed
Nov 25, 2021
| @@ -69,7 +69,7 @@ instance (Typeable t, FromCBOR (Annotator t)) => FromCBOR (Annotator (MemoBytes | |||
| (Annotator getT, Annotator getBytes) <- withSlice fromCBOR | |||
| pure (Annotator (\fullbytes -> Memo (getT fullbytes) (toShort (toStrict (getBytes fullbytes))))) | |||
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| instance Eq t => Eq (MemoBytes t) where (Memo x _) == (Memo y _) = x == y | |||
| instance Eq (MemoBytes t) where (Memo _ x) == (Memo _ y) = x == y | |||
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Not comparing both constituents of Memo implicitly supposes that if two binary representations are equal, then their Haskell representation must be equal a well; is there anything truly enforcing that? While it sounds plausible in my head, is it really the case 🤔 ?
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This was what I was justifying above:
serialise x == serialise y
deserialise (serialise x) == deserialise (serialise y)
x == y
The property that deserialise is left-inverse to serialise is a required property for any serialisable type (as opposed to the other way around!). And we have tests verifying this property for serialised types.
These are not needed now that `MemoBytes` is always comparable.
redxaxder
approved these changes
Nov 25, 2021
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Two instances of a memoised type are equal iff they have the same bytes.
Justification:
if we have
serialise x == serialise ydeserialise (serialise x) == deserialise (serialise y)x == yEqshould be compatible with hashing (e.g. ifa == b, thenhash a == hash b). Currently, this is broken. If I serialise aTxBodyRawin two different valid ways (as is perfectly allowed),then the resulting
TxBodywill currently compare as equal but hashto a different value.
bytes are an integral part of the type, and must be considered in
comparing for equality.