Is it the correct way to check for specializations?

General Usage
1

Following the last paragraph at docs, I want to check if julia is generating specialized version for some of my code. Testing against some built-in functions, first define

function get_sp(wfx)
    for mi in wfx.specializations
        if !isnothing(mi)
            println(mi)
            println("-"^9)
        end
    end
end

Then, running

exp(1)
exp(rand(3, 3))
exp(rand(Float32, 3, 3))
exp(rand(ComplexF64, 3, 3))

get_sp(@which exp(rand(3, 3)))

shows

MethodInstance for exp(::Matrix{Float64})
---------
MethodInstance for exp(::Matrix{ComplexF64})
---------
MethodInstance for exp(::Matrix{Float32})
---------

while

sin(1)
sin(1.2)
sin(0.3im)
get_sp(@which sin(.3im))

just gives

MethodInstance for sin(::ComplexF64)
---------

Is it then correct to say that, for exp(m) specialzied versions are generated for different types of matrices while no specialized version is generated for different number types to sin(x)?

Thanks!

2

which returns just one method, and gives the specializations of that one method. In this case, I think you want to check all methods of sin to see what specializations each of them had

1 Like
3

Beware, though, that which can create specializations. A fairly safe approach is to use the MethodAnalysis package:

julia> using MethodAnalysis

julia> methodinstances(sin)
Core.MethodInstance[]

julia> sin(3)
0.1411200080598672

julia> sin(3.5)
-0.35078322768961984

julia> methodinstances(sin)
2-element Vector{Core.MethodInstance}:
 MethodInstance for sin(::Float64)
 MethodInstance for sin(::Int64)

julia> @which sin(3)
sin(x::Real)
     @ Base.Math math.jl:1488

julia> @which sin(3.5)
sin(x::T) where T<:Union{Float32, Float64}
     @ Base.Math special/trig.jl:29

So you can see it aggregated the specializations across multiple methods, once those specializations were created.

If you start a fresh session and reverse the order, calling @which before the first invocation of methodinstances, you’ll see those specializations already exist—they were created by @which.

3 Likes
4

Thanks for the package!

5

if which is to be avoided when looking for specializations, then the documentation needs to be updated. see the last sentence in the section linked to above.