You're getting confused by what the value of an assignment expression is. This
is what I think you're hoping your code does (but it doesn't do that):
function makestring(fun)
A::Array{ASCIIString} = [fun(i) for i = 1:3]
return A
end
-- John
On Oct 29, 2014, at 10:57 AM, Zenna Tavares <zennatava...@gmail.com> wrote:
> Ok, but then why does it return the Vector{Any}. The following two functions
> give different results
>
> function makestring(fun)
> A::Array{ASCIIString} = [fun(i) for i = 1:3]
> end
>
>
> function makestring(fun)
> A::Array{ASCIIString} = [fun(i) for i = 1:3]
> convert(Array{ASCIIString},A)
> end
>
> I thought maybe it was a weird REPL thing, but it's not.
>
> On Wednesday, October 29, 2014 1:52:01 PM UTC-4, John Myles White wrote:
> I'm pretty sure this sort of thing always works since type declarations on
> variables behave like convert calls:
>
> julia> function foo()
> a::Int64 = 0x01
> return a
> end
> foo (generic function with 1 method)
>
> julia> foo()
> 1
>
> -- John
>
> On Oct 29, 2014, at 10:50 AM, Zenna Tavares <zennat...@gmail.com> wrote:
>
> > Also, the following runs but still returns a Vector{Any}. How is this
> > possible?
> >
> > function makestring(fun)
> > A::Array{ASCIIString} = [fun(i) for i = 1:3]
> > end
> >
> >
> > On Wednesday, October 29, 2014 1:45:51 PM UTC-4, Zenna Tavares wrote:
> > As shown in the following example, I am getting differently typed arrays
> > depending on where I use the list comprehension.
> >
> > f(i) = "$i"
> > A = [f(i) for i = 1:3]
> > function makestring(fun)
> > A = [fun(i) for i = 1:3]
> > end
> > B = makestring(f)
> >
> > In this example A has type Vector{ASCIIString} while B has type
> > Vector{Any}. What gives? And is there a workaround such that we get a more
> > specific type?
> >
> > I understand there are some open issues related to this, but I am not sure
> > if this case comes under what's already been discussed.
> >
> >
>
