Julia: Vararg version of hypot does overflow and underflow on master

[giordano]

The point of the hypot function is to avoid underflow and overflow, however the implementation of the vararg method introduced with PR #25571 defeats this purpose:

julia> hypot(1e200, 0, 0)
Inf

julia> hypot(1e-300, 0, 0)
0.0

Instead in Julia 0.6 it gives the correct result:

ulia> hypot(1e200, 0, 0)
1.0e200

julia> hypot(1e-300, 0, 0)
1.0e-300

Edit: I've just seen this was already pointed out in https://github.com/JuliaLang/julia/pull/25571#discussion_r162442508

[fredrikekre]

[stevengj]

Maybe just define:

hypot(x::Number...) = vecnorm(x)

[simonbyrne]

@stevengj I think the problem is that vecnorm is part of LinAlg.

One alternative would be

hypot(x,y,z...) = hypot(hypot(x,y), z...)

[stevengj]

That would be a performance trap, in that it would be much slower than expected.

We can also just reimplement the relevant part of vecnorm: it doesn’t take more than a few lines to do the right thing.

[stevengj]

The following cases give even worse failures due to integer wraparound:

julia> i = typemax(Int)
9223372036854775807

julia> hypot(i,i,i)  # the result is √3 !!
1.7320508075688772

julia> i*√3
1.5975348984942514e19

julia> hypot(10^17, 10^17, 10^17)  # 3*abs2(10^17) is negative!
ERROR: DomainError with -6.437707461859213e18:

There should definitely be a test for the i = typemax(Int) case in any solution PR.