Hi! I have an issue with allocations using the `mul!`

function and a closure:

```
using LinearAlgebra, Test
struct operator{T}
prod!
end
function operator(A :: AbstractArray{T}) where T
prod! = (res, u, α, β) -> mul!(res, A, u, α, β)
return operator{T}(prod!)
end
import LinearAlgebra.mul!
function mul!(res::Vector{T}, op::operator{T}, u::Vector{T}, α::T, β::T) where T
op.prod!(res, u, α, β)
end
function mul!(res::Vector{T}, op::operator{T}, u::Vector{T}) where T
op.prod!(res, u, one(T), zero(T))
end
function mul2!(res::Vector{T}, op::operator{T}, u::Vector{T}) where T
op.prod!(res, u, 2*one(T), 2*one(T))
end
function test_allocs()
n = 10
A = rand(n, n)
A_op1 = operator(A)
res1, u1, α1, β1 = rand(n), rand(n), 2.0, 2.0
### compile
A_op1.prod!(res1, u1, β1, α1)
mul!(res1, A_op1, u1, β1, α1)
mul!(res1, A, u1, α1, β1)
mul2!(res1, A_op1, u1)
### tests allocs
allocs1 = @allocated A_op1.prod!(res1, u1, α1, β1)
@test allocs1 == 16
allocs2 = @allocated mul!(res1, A_op1, u1, α1, β1)
@test allocs2 == 32
allocs3 = @allocated mul!(res1, A, u1, α1, β1) # LinearAlgebra mul! function for matrices
@test allocs3 == 0
allocs4 = @allocated mul!(res1, A_op1, u1)
@test allocs4 == 0
allocs5 = @allocated mul2!(res1, A_op1, u1)
@test allocs5 == 0
end
test_allocs()
```

All tests passed but I do not know how to remove the allocations for the first and second tests with my type `operator{T}`

. Maybe it comes from `α`

and `β`

because when I removed them with the 3-args `mul!`

and `mul2!`

the allocations disappear, but I could not find a way to solve this problem. Does anybody know how I could fix this?