F#: how to evaluate a “seq” to get all its values eagerly?

问题

We know that in F#, seq is lazy evaluated. My question is, if I have a seq with limited number of values, how to convert it into some data type that contains all its value evaluated?

> seq { for i in 1 .. 10 do yield i * i };;
val it : seq<int> = seq [1; 4; 9; 16; ...]

Thanks a lot.

回答1:

The answer from @Carsten is correct: you can use Seq.toArray or Seq.toList if you wish to convert lazily evaluated sequences to lists or arrays. Don't use these function to force evaluation, though.

The most common reason people tend to ask about this is because they have a projection that involves side effects, and they want to force evaluation. Take this example, where one wishes to print the values to the console:

let lazySeq = seq { for i in 1 .. 10 do yield i * i }
let nothingHappens = lazySeq |> Seq.map (printfn "%i")

The problem is that when you evaluate these two expressions, nothing happens:

> 

val lazySeq : seq<int>
val nothingHappens : seq<unit>

Because nothingHappens is a lazily evaluated sequence, no side-effects occur from the map.

People often resort to Seq.toList or Seq.toArray in order to force evaluation:

> nothingHappens |> Seq.toList;;
1
4
9
16
25
36
49
64
81
100
val it : unit list =
  [null; null; null; null; null; null; null; null; null; null]

While this works, it's not particularly idiomatic; it produces a weird return type: unit list.

A more idiomatic solution is to use Seq.iter:

> lazySeq |> Seq.iter (printfn "%i");;
1
4
9
16
25
36
49
64
81
100
val it : unit = ()

As you can see, this forces evaluation, but has the more sane return type unit.

回答2:

use Seq.toArray (for an array) or Seq.toList (for an list) ;)

there are plenty more - just choose ;)

example:

> seq { for i in 1 .. 10 do yield i * i } |> Seq.toArray;;
val it : int [] = [|1; 4; 9; 16; 25; 36; 49; 64; 81; 100|]

来源：https://stackoverflow.com/questions/35169269/f-how-to-evaluate-a-seq-to-get-all-its-values-eagerly

标签

evaluation

seq

eager