Array{Int,3}() should create an empty 3d array

[eschnett]

Array{Int,1}() and Array{Int,2}() create empty 1d- and 2d-arrays, respectively. This could also be implemented for all higher dimensions.

[rofinn]

+1 I ran into this last week and had to do Array(Int, 2, 3, 4) instead.

[yuyichao]

Note that high dimension arrays cannot be resized so it's unclear what is Array{T,N}() with N >= 2 good for. On the other hand, I think it's useful to allow Array{Int,3}(2, 3, 4) in additional to Array{Int}(2, 3, 4) and Array(Int, 2, 3, 4).

@Rory-Finnegan If you want Array(Int, 2, 3, 4) an empty 3-d array won't be useful for you.

[rofinn]

@yuyichao Sorry, my initial use case was an empty 3-d array, but when I couldn't figure out how to do that I just preallocated a large one and used sub once I had filled it in (not the best solution, but meh). In either case, it did surprise me that Array{Int,3}() and Array{Int,3}(2,3,4) weren't valid constructors.

[JeffBezanson]

Honestly I'm not a fan of Array{Int,2}() making a 0x0 array. I find that both non-obvious and useless. +1 to the versions that accept size arguments though.

[GunnarFarneback]

Note that high dimension arrays cannot be resized so it's unclear what is Array{T,N}() with N >= 2 good for.

Type-stable placeholders.

[nalimilan]

+1, at the very least to ensure consistency.

[stevengj]

Consistency with what?

[KristofferC]

With N=1 and N=2?

[StefanKarpinski]

I'm not sure why this works for N=1 and N=2 in the first place. That seems like a misfeature to me. It's as if we've decided that the default size of dimensions is zero but only if we don't specify any dimensions. What?!?! That strikes me as absurd and I'm somewhat tempted to find out when we started supporting this. (It's probably my fault, 😬 .) For N=0 it makes sense since there are no dimensions to specify.

[StefanKarpinski]

The constructors with specified dimensions already all work:

Array{Int,3}((2,3,4))
Array{Int}((2,3,4))
Array{Int,3}(2,3,4)
Array{Int}(2,3,4)

[StefanKarpinski]

The N=1 case actually kind of makes sense since we can at least grow those.

[martinholters]

The problem I have with the N==1 case is that it makes one think that if Vector{Int}() is equivalent to Int[], then Vector{Int}(42) is equivalent to Int[42].

I agree that as a beginner, Vector{Int}() is one of the first things to try to create an empty vector. But you better learn using [] quickly anyway, as you'll need it for non-empty vectors of all kinds (concatenation, comprehension).

[StefanKarpinski]

I'm more ok with Vector{Int}() and Vector() because zero seems like a sensible default for the length of a growable vector, but I agree that there's a solid argument to be made for its removal.

[KristofferC]

Closed by #20330