I think I do not quite understand how F# infers types in sequence expressions and why types are not correctly recognized even if I specify the type of the elements directly from "seq".
In the following F# code we have a base class A and two derived classes, B and C:
type A(x) =
member a.X = x
type B(x) =
inherit A(x)
type C(x) =
inherit A(x)
If I try to "yield" their instances in a simple sequence expressions, I get two errors:
// Doesn't work, but it makes sense.
let testSeq = seq {
yield A(0)
yield B(1) // Error, expected type: A
yield C(2) // Error, expected type: A
}
That can make sense, since it may not be so trivial to infer "common" types (interfaces, I think, can make that work far harder). However, those errors can be fixed with a safe cast:
// Works fine :)
let testSeqWithCast = seq {
yield A(0)
yield B(1) :> A
yield C(2) :> A
}
What if I do not want to use casts? I tried to specify the sequence type directly from "seq", but things do not seem to work:
// Should work, I think...
let testGenSeq = seq<A> {
yield A(0)
yield B(1) // Error, expected type: A
yield C(2)
}
So, my question is: is there a way to avoid casts? If not, is there a reason why even specifying the type doesn't make the code work?
I tried digging through following links:
http://msdn.microsoft.com/en-us/library/dd233209.aspx http://lorgonblog.wordpress.com/2009/10/25/overview-of-type-inference-in-f/
But I found nothing useful...
Thank you in advance for any kind of answer you can give :)
In order to understand the cause of your confusion you should not go anywhere further, than the first statement of the link you referred to :
A sequence is a logical series of elements all of one type.
You can return a sequence of only one, the same type like seq<A>
, or seq<obj>
. The OOP-ish fact that types B
and C
are inherited from A
is not relevant. The following may help: all your instances are also inherited from obj
, but in order to make from them a seq<obj>
you should explicitly cast:
// Works fine
let testSeq = seq<obj> {
yield A(0) :> obj
yield B(1) :> obj
yield C(2) :> obj
}
or just box
them like below:
// Works fine too
let testSeq = seq {
yield box (A(0))
yield box (B(1))
yield box (C(2))
}
EDIT: For understanding the reasoning behind explicit casting in F# the following (simplistic) consideration may help. Type inference does not do guessing; unless it can derive seq
type deterministically, or have it explicitly declared, it will complain.
If you just do
let testSeq = seq {
yield A(0)
yield B(1)
yield C(2)
}
compiler is presented with indeterminism - testSeq
can be either seq<A>
, or seq<obj>
, so it complains. When you do
let testSeq = seq {
yield A(0)
yield upcast B(1)
yield upcast C(2)
}
it infers testSeq
as seq<A>
based on type of the first member and upcasts B and C to A
without complaining. Similarly, if you do
let testSeq = seq {
yield box A(0)
yield upcast B(1)
yield upcast C(2)
}
it will infer testSeq
as seq<obj>
based on the type of the first member upcasting this time second and third members to obj
, not A
.