Inference regression for empty array

[nalimilan]

DataFrames formula code hits an inference regression with an empty array. This is problematic since ! is now only defined for AbstractArray{Bool} (which might be an issue on its own).

function f()
    x = [1]
    x = x[!(x .== 1)]
    x .== 0
end

function g()
    x = Any[1]
    x = x[!(x .== 1)]
    x .== 0
end

function h()
    x = Any[1]
    x .== 0
end

On Julia 0.5.0:

julia> f()
0-element BitArray{1}

julia> g()
0-element BitArray{1}

julia> h()
1-element BitArray{1}:
 false

On master:

julia> f()
0-element BitArray{1}

julia> g()
0-element Array{Any,1}

julia> h()
1-element BitArray{1}:
 false

[Sacha0]

(Tangentially, perhaps we should deprecate ! vectorized over AbstractArray{Bool}? Also, !.[true] yields ERROR: syntax: invalid identifier name "." (#20039). Best!)

[pabloferz]

In the meantime, if you need a workaround, you can do

iseq(x, y)::Bool = x == y

function g()
    x = Any[1]
    x = x[(!).(iseq.(x, 1))]
    iseq.(x, 0)
end

[KristofferC]

The original code is deprecated for broadcasted ! which works fine.

[tkelman]

works fine

How so? I still see

function g()
    x = Any[1]
    x = x[(!).(x .== 1)]
    x .== 0
end
g()

return a 0-element Array{Any,1}

also with

julia> function g2()
           x = Any[1]
           x = x[.!(x .== 1)]
           x .== 0
       end
g2 (generic function with 1 method)

julia> g2()
0-element Array{Any,1}

[KristofferC]

Yeah, I goofed here.

[vtjnash]

we can simplify the original question to simply:

julia> [] .== 0
0-element Array{Any,1}

Which further simplifies to:

Core.Inference.return_type(==, (Any, Int)) === Bool

However, we haven't typically been willing to commit to forcing == to always return Bool, so this isn't expected to work.