Why does my hash function return a different value to the built in hash function?

Jon_Barker · March 16, 2022, 2:18pm

I copied the hash function code from here:

JuliaLang/julia/blob/v1.7.0/base/hashing.jl

# This file is a part of Julia. License is MIT: https://julialang.org/license

## hashing a single value ##

"""
    hash(x[, h::UInt]) -> UInt

Compute an integer hash code such that `isequal(x,y)` implies `hash(x)==hash(y)`. The
optional second argument `h` is a hash code to be mixed with the result.

New types should implement the 2-argument form, typically by calling the 2-argument `hash`
method recursively in order to mix hashes of the contents with each other (and with `h`).
Typically, any type that implements `hash` should also implement its own [`==`](@ref) (hence
[`isequal`](@ref)) to guarantee the property mentioned above. Types supporting subtraction
(operator `-`) should also implement [`widen`](@ref), which is required to hash
values inside heterogeneous arrays.

See also: [`objectid`](@ref), [`Dict`](@ref), [`Set`](@ref).
"""
hash(x::Any) = hash(x, zero(UInt))

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And ran it under J1.7 but I get different results. Why?

Jon_Barker · March 16, 2022, 2:24pm

Your second line - that should be:

hash2(x::Any) = hash2(x, zero(UInt))

?

lawless-m · March 16, 2022, 2:27pm

Ok, let me rephrase that
I get the same

julia> hash2((1,2))
0xc04cd80fdd457cea

Jon_Barker · March 16, 2022, 2:30pm

Still no clue why I don’t get that then:

hash2(x::Any) = hash2(x, zero(UInt))
hash2(@nospecialize(x), h::UInt) = Base.hash_uint(3h - objectid(x))
hash((1,2))==hash2((1,2))

Jon_Barker · March 16, 2022, 2:31pm

Ah i saw you edited your post - OK good I’m not going crazy.
Anyone shed some light on this?

pfitzseb · March 16, 2022, 2:32pm

There’s a special hash function for tuples:

julia> @which hash((1,2), zero(UInt))
hash(t::Tuple, h::UInt64) in Base at tuple.jl:417

lawless-m · March 16, 2022, 2:32pm

I even took hashing.jl, replaced all the hash( with hash2( and got the same result

Jon_Barker · March 16, 2022, 2:33pm

@which
Everytime I post here I learn of a new helpful macro. Thanks!