I typed methods(√) and got, among other entries, this: sqrt(A::Union{DenseArray{#s627,2}, Base.ReinterpretArray{#s627,2,S,A} where S where A<:Union{SubArray{T,N,A,I,true} where I<:Union{Tuple{Vararg{Real,N} where N}, Tuple{AbstractUnitRange,Vararg{Any,N} where N}} where A<:DenseArray where N where…

Which Julia version do you have? On my v1.3, sqrt and √ both look fine: julia> methods(sqrt) # 19 methods for generic function "sqrt": [1] sqrt(a::Float16) in Base.Math at math.jl:1085 [2] sqrt(a::Complex{Float16}) in Base.Math at math.jl:1086 [3] sqrt(::Missing) in Base.Math at math.jl:1138 [4] sq…

Your methods 10 and 11 are just as long! Judging from that line, seems it stems from LinearAlgebra and the truly humongous number of where classifieres suggests this is just an expanded chain of type definitions.

True, haha. That does not seem to be a bother, though.

Why is the strange type signature of sqrt so long?

New to Julia

kristoffer.carlsson December 21, 2019, 9:00am 5

It is because of StridedArray being a union of a large number of complicated types.

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Why is the strange type signature of sqrt so long?

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