Hash instability in Julia composite types

Question

In Julia, composite types with at least one field that have the same hash values ​​for different values. This means that composite types do not work correctly if you use them as dictionary keys or anything else that depends on a hashed value. This behavior is incompatible with the behavior of other types, for example Vector {Int}.

More specific,

vectors of unconsolidated types that are different objects but have the same hash value with the same value:

julia> hash([1,2,3])==hash([1,2,3]) true 

hash fieldless composite types with the same value:

 julia> type A end julia> hash(A())==hash(A()) true 

composite types with at least one field hash for different values, if they are different objects with the same value:

 julia> type B b::Int end julia> hash(B(1))==hash(B(1)) false 

however, the same object retains its hash even if the base values ​​change:

 julia> b=B(1) julia> hash(b) 0x0b2c67452351ff52 julia> bb=2; julia> hash(b) 0x0b2c67452351ff52 

this is incompatible with the behavior of vectors (if you change the item, hash changes):

 julia> a = [1,2,3]; julia> hash(a) 0xd468fb40d24a17cf julia> a[1]=2; julia> hash(a) 0x777c61a790f5843f 

this problem is missing for immutable types:

 julia> immutable C c::Int end julia> hash(C(1))==hash(C(1)) true 

Is there something fundamental that governs this behavior in terms of language design? Are there any plans to fix or fix this behavior?

Answer 1

I am not a Julia language developer, but I say that this behavior is not surprising when you compare volatile and unchanging values. Your type B changed: it is not clear that two instances of it, even if they have the same value for field B , should be considered equal. If this seems to be the case, you can implement a hash function for it. But in general, mutable objects have independent identities. Immutable entities, such as C , cannot be distinguished from each other, so it is reasonable for them to obey structural hashing.

Two bank accounts with $ 5 in them are not identical and probably should not be hashed with the same number. But two copies of $ 5 cannot be distinguished from each other.