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.

Unexpectedly long method signature in "Closest candidates" list

Topic		Replies	Views
Unexpectedly long method signature in "Closest candidates" list General Usage question	4	559	December 21, 2019
Create a more unified syntax for requesting a specific return type of functions? General Usage functions	18	357	November 1, 2024
Some doubts about types New to Julia question	9	1994	August 31, 2019
Should Julia have rsqrt? And e.g. support RSQRTSS Internals & Design	12	1210	February 20, 2025
Why does scientific notation break the range function? New to Julia	95	4582	April 29, 2022

Why is the strange type signature of sqrt so long?

Related topics