check that hashing of types does not foreigncall (jl_type_hash is concrete evaluated) (#58401)

(cherry picked from commit 86a3e1a)

Author: KristofferC

base/hashing.jl

@@ -29,7 +29,8 @@ hash(x::Any) = hash(x, zero(UInt))
 hash(w::WeakRef, h::UInt) = hash(w.value, h)
 
 # Types can't be deleted, so marking as total allows the compiler to look up the hash
-hash(T::Type, h::UInt) = hash_uint(3h - @assume_effects :total ccall(:jl_type_hash, UInt, (Any,), T))
+@noinline _jl_type_hash(T::Type) = @assume_effects :total ccall(:jl_type_hash, UInt, (Any,), T)
+hash(T::Type, h::UInt) = hash_uint(3h - _jl_type_hash(T))
 
 ## hashing general objects ##
 

test/hashing.jl

@@ -303,4 +303,10 @@ struct AUnionParam{T<:Union{Nothing,Float32,Float64}} end
     @test hash(5//3) == hash(big(5)//3)
 end
 
-@test Core.Compiler.is_foldable_nothrow(Base.infer_effects(hash, Tuple{Type{Int}, UInt}))
+@testset "concrete eval type hash" begin
+    @test Core.Compiler.is_foldable_nothrow(Base.infer_effects(hash, Tuple{Type{Int}, UInt}))
+
+    f(h...) = hash(Char, h...);
+    src = only(code_typed(f, Tuple{UInt}))[1]
+    @test count(stmt -> Meta.isexpr(stmt, :foreigncall), src.code) == 0
+end