optimize undefined variables better

[simonp0420]

As shown below, fun1 takes 10x as long and allocates a large amount of memory, compared to fun2. fun2 differs only by initializing the local variable i.

julia> x = rand(10000000);

julia> function fun1(x::Vector{Float64})
           local i::Int
           for i = 1:length(x)
               if x[i] == 0.0
                   break
               end
           end
           return i-1
       end
fun1 (generic function with 1 method)

julia> function fun2(x::Vector{Float64})
           local i::Int = 0
           for i = 1:length(x)
               if x[i] == 0.0
                   break
               end
           end
           return i-1
       end
fun2 (generic function with 1 method)

julia> fun1(x); fun2(x); # warmup

julia> @time fun1(x)
elapsed time: 0.199774304 seconds (159998472 bytes allocated)
9999999

julia> @time fun2(x)
elapsed time: 0.020258781 seconds (64 bytes allocated)
9999999

[JeffBezanson]

The real issue is that this code doesn't handle length(x)==0. Arguably we should give a compiler warning for that.

[StefanKarpinski]

I think it would be good to issue a warning if a function can return an undefined value. I've also been thinking that it might be good to integrate bits of TypeCheck into Base so that it's easier to use.

[IainNZ]

Definitely a good idea Stefan, anything to save people from them selves.

You could maybe even go far as to say that its easier to make performance related mistakes in Julia for newbies (or at least, it seems like from reading the mailing lists that in terms of speed, bad julia < bad python/matlab < good python/matlab < good julia).

[StefanKarpinski]

Another thought is that it could be an error to return an undefined value. That would allow this code to run at full speed and simply error out if x is empty.

[simonp0420]

+1. A perhaps more reasonable example where this might be useful, but that currently is slow:

function fun1(x::Vector{Float64})
    @assert !isempty(x)
    local i::Int
    for i = 1:length(x)
        if x[i] == 0.0
            break
        end
    end
    return i-1
end

[JeffBezanson]

@StefanKarpinski it doesn't need to specifically be an error to return an undefined value, because you get an error as soon as you access the undefined variable. The code already errors out if x is empty. The only reason it doesn't run at full speed is that we box i in order to make it "nullable".

[JeffBezanson]

Fixed by #18732.