Hi,
why does not this code work?
type Test() =
static member func (a: seq<'a seq>) = 5.
let a = [[4.]]
Test.func(a)
It gives following error:
The type 'float list list' is not compatible with the type 'seq<seq<'a>>'
Hi,
why does not this code work?
type Test() =
static member func (a: seq<'a seq>) = 5.
let a = [[4.]]
Test.func(a)
It gives following error:
The type 'float list list' is not compatible with the type 'seq<seq<'a>>'
The error message describes the problem -- in F#, list<list<'a>>
isn't compatible with seq<seq<'a>>
.
The upcast
function helps get around this, by making a
into a list<seq<float>>
, which is then compatible with seq<seq<float>>
:
let a = [upcast [4.]]
Test.func(a)
Edit: You can make func
more flexible in the types it accepts. The original accepts only sequences of seq<'a>
. Even though list<'a>
implements seq<'a>
, the types aren't identical, and the compiler gives you an error.
However, you can modify func
to accept sequences of any type, as long as that type implements seq<'a>
, by writing the inner type as #seq
:
type Test() =
static member func (a: seq<#seq<'a>>) = 5.
let a = [[4.]]
Test.func(a) // works
Change your code to
type Test() =
static member func (a: seq<#seq<'a>>) = 5.
let a = [[4.]]
Test.func(a)
The trick is in the type of a. You need to explicitly allow the outer seq to hold instances of seq<'a> and subtypes of seq<'a>. Using the # symbol enables this.