Copy! in interpolation does not take into account scalars

[bertini97]

Describe the bug 🐞

> a = LinearInterpolation([1.0, 2.0], [1.0, 2.0])
LinearInterpolation{Vector{Float64}, Vector{Float64}}([1.0, 2.0], [1.0, 2.0], false)

> b = [1.0]
1-element Vector{Float64}:
 1.0

> a(b, 1.1)
1-element Vector{Float64}:
 1.1

> a(b, 1.0)
ERROR: MethodError: no method matching copy!(::Vector{Float64}, ::Float64)

This is because of the copy! when the tval is the first one:

SciMLBase.jl/src/interpolation.jl

Line 291 in eaebe6e

copy!(out, u[k])

Expected behavior
Expected behavior is that it should be interpolated properly. I don't know if this is intended. It doesn't look like it to me. Cheers.

[bertini97]

A solution that works is doing what is done below for all other values:

out .= u[k]

[ChrisRackauckas]

Is this a DataInterpolations.jl issue? This is for DataInterpolations.LinearInterpolation?

[bertini97]

I would say no, I don't have it installed. I imported SciMLBase's interpolations.jl

[ChrisRackauckas]

Oh I see what you're saying. The inplace form of interpolation just cannot support scalars. That is just a fundamental Julia thing and cannot be solved by a package (nor does it really make sense? There is no preallocating for a scalar)

[bertini97]

Sorry, I improperly called them scalars. But they are in fact a 1D array with one element. The interpolation should work. And it does if you do not select a boundary value. If you select a boundary value it will use copy instead of broadcast, for reasons I can't tell right now. But this looks like a bug, I urge you to consider reopening.