And if you want to reduce memory allocations and your fields are all of the
same type:
immutable Foo{T}
a::T
b::T
c::T
end
# this always works also for mutables / varying fieldtypes:
typevec(a) = [a.(x) for x in names(a)]
# create initial view
typevec!(foo::Foo) = pointer_to_array(
convert(Ptr{typeof(foo.a)},
pointer_from_objref(foo))+sizeof(typeof(foo.a)), 3)
# point to a different obj, reusing the view array
function typevec!(a, view)
p = convert(Ptr{eltype(view)}, pointer_from_objref(view))
unsafe_store!(p, pointer_from_objref(a)+sizeof(eltype(view)), 2)
view
end
foo = Foo(1,2,3)
foo2 = Foo(10,20,20)
@show @time typevec(foo)
@show @time view = typevec!(foo)
@show @time typevec!(foo2,view)
Just be careful that foo does not go out of scope while you use the result of
"typevec!"
Am 05.03.2015 um 11:00 schrieb Simon Danisch <[email protected]>:
> easiest would probably be: [foo.(i) for i=1:length(names(Foo))]
>
>
> Am Donnerstag, 5. März 2015 10:51:36 UTC+1 schrieb Tamas Papp:
> If I have a composite type
>
> type Foo
> a
> b
> c
> end
>
> I can initialize it with a vector like this:
>
> x = Foo([1,2,3]...)
>
> But how can I get the fields as a vector, eg [x.a,x.b,x.c] without
> enumerating them? convert(x, Vector) does not work.
>
> I need this to go back and forth between the type and the vector
> representation of a set of parameters when using NLopt.
>
> Best,
>
> Tamas
