I noticed a strange behavior when using constructors of parametric types. If a constructor does not have constraints on type parameters, then both type inference and providing explicit type argument work: struct Foo{T} val :: T Foo(x :: T) where T = new{T}(x) end julia> Foo(3) Foo{Int64}(3) …

Ways of using constructor of a parametric type

julbinb September 25, 2017, 1:08pm 3

Thank you for reply!

This is 0.7.0.

1 Like

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