Trying to understand where all the memory allocations in a call to "notify" are coming from.

[Amit Murthy]

Consider the following code:

julia> function foo()
           c=Condition()
           for i in 1:10^6
               notify(c)
           end
       end
foo (generic function with 1 method)
 
julia> foo()
 
julia> @time foo()
 158.151 milliseconds (2000 k allocations: 122 MB, 4.75% gc time)

That is a lot of memory allocations.

`notify` in task.jl is defined as 

function notify(c::Condition, arg::ANY=nothing; all=true, error=false)
    if all
        for t in c.waitq
            schedule(t, arg, error=error)
        end
        empty!(c.waitq)
    elseif !isempty(c.waitq)
        t = shift!(c.waitq)
        schedule(t, arg, error=error)
    end
    nothing
end

and in this case c.waitq is always empty.

If I specify a keyword arg, for example, `notify(c; all=false)`, number of mem allocations goes up 4 times.

julia> function foo()
           c=Condition()
           for i in 1:10^6
               notify(c; all=false)
           end
       end
foo (generic function with 1 method)
 
julia> foo()
 
julia> @time foo()
   2.445 seconds      (8000 k allocations: 381 MB, 2.73% gc time)

Considering the `notify` call:

julia> const ct=Condition()
Condition(Any[])

julia> @code_typed notify(ct)
1-element Array{Any,1}:
 :($(Expr(:lambda, Any[:c], Any[Any[Any[:c,Condition,0]],Any[],Any[Union{},Union{},Union{},Union{},Union{},Union{},Union{},Union{},Array{Any,1},Array{Any,1},Int64],Any[]], :(begin $(Exp
r(:line, 236, symbol("task.jl"), symbol("")))
        GenSym(8) = (top(ccall))(:jl_alloc_array_1d,(top(apply_type))(Array,Any,1)::Type{Array{Any,1}},(top(svec))(Any,Int)::SimpleVector,Array{Any,1},0,0,0)::Array{Any,1}
        GenSym(9) = GenSym(8)
        return __notify#26__(GenSym(9),c::Condition)::Void
    end::Void))))

Why is there a `ccall` and `jl_alloc_array_1d`? Is this due to pre-compilation?  What exactly is happening here?

[Yichao Yu]

These are coming from the packing of keyword arguments and is exactly
why they are very slow.

>
>

[Amit Murthy]

OK. Also noticed that,

function bar(x;a=1, b=2)

    x+a+b

end

is OK when invoked as bar(1), but slow when invoked as bar(1; a=1)

but

function bar(x, y=1; a=1, b=2)

    x+a+b

end

is slow even when invoked as bar(1)

[Yichao Yu]

On Sat, Jul 4, 2015 at 12:15 PM, Amit Murthy <amit....@gmail.com> wrote:
> OK. Also noticed that,
>
> function bar(x;a=1, b=2)
> x+a+b
> end
>
> is OK when invoked as bar(1), but slow when invoked as bar(1; a=1)
>
> but
>
> function bar(x, y=1; a=1, b=2)
> x+a+b
> end
>
> is slow even when invoked as bar(1)

It's fast if invoked as bar(1, 1). This one can probably be fixed
without too much effort.

[Andreas Noack]

Maybe this should be filed as an issue such that we don't lose track of it.