Passing F# function to IEnumerable.Where vs IEnumerable.All

Question

Given the following:

open System.Linq

let even n = n % 2 = 0

let seqA = seq { 0..2..10 }

this is a valid expression:

seqA.Where(even)

but this is not:

seqA.All(even)

Why is passing even to Where allowed but not to All?

My guess is you've discovered a bug. I would send this to [email protected]. — Daniel
This seems to be another case of this. As a workaround, seqA.All(System.Func<_,_>(even)) should work, or seqA.All(fun x -> even x). — Christoph Rüegg
@Daniel fsbugs requested that I open an issue on their codeplex site. Here it is. — dharmatech
even |> seqA.All does work (on mono). Just to make things more interesting. — albertjan

Reid Evans Reid Evans · Accepted Answer · 2015-05-29T15:43:10

While there may be a bug there, I think the better approach would be to use the Seq higher order functions when dealing with IEnumerable<T> in F# rather than Linq

let even n = n % 2 = 0
let seqA = seq { 0..2..10 }

seqA |> Seq.filter even
//val it : seq<int> = seq [0; 2; 4; 6; ...]

seqA |> Seq.forall even
//val it : bool = true

Passing F# function to IEnumerable.Where vs IEnumerable.All

1 Answers