Two type composite struct construction

New to Julia
1

Hi.
I have the following problem/question. I have a struct

struct foo{T<:Integer, U<:Real} <: AbstractGraph{T}
  bar1::T
  bar2::U
end

Now, I want to create an instance of foo, where I explicitly define the type U but infer the type T from the arguments. I tried

function foo{U}(...) where {U<:Real}
  # do some stuff here and get T
  foo{T,U}(bar1, bar2)
end

But this errored with asking for T. Now, what is the right way to construct foo with only passing U as a type?

EDIT1: See the marked answer and have a look at this thread as well: https://discourse.julialang.org/t/how-to-correctly-define-and-call-templated-parametric-methods-using-the-new-where-syntax

2

Something like that?

function foo(::Type{U}, bar2) where {U<:Real}
    # do some stuff here and get T
    T = typeof(bar2)
    bar1 = zero(U)
    foo{T,U}(bar1, bar2)
end
3

Hi @Henrique_Becker,

thanks, that was what I was looking for!

4

I noticed just now that this is basically the first doubt I asked in Discourse.

So the inquirer is now the replier. The cycle begins anew.

5

Very good! Reading through your linked thread leaves me with another question. What is the difference between below definitions of foo

foo(::Type{U}) where {U<:Real}

and

foo(U::Type)

besides that U can by any type in the second example und only a subtype of real in the first?

6

There is no difference. Actually, if you define foo at the REPL, you will see that after the second definition, there is still only one method!

julia> foo(::Type{U}) where U = 1
foo (generic function with 1 method)

julia> foo(U::Type) = 100
foo (generic function with 1 method)
7

That’s interesting. So, why are there two different syntax’?

8

As in your example, writing Type{T} where T gains you access to the type parameter either for dispatch purposes (T <: Real) or to use later inside your function.

9

that makes sense! Great, thanks!