本文介绍了将 F# 函数传递给 IEnumerable.Where 与 IEnumerable.All的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
鉴于以下内容:
open System.Linq let even n = n % 2 = 0 let seqA = seq { 0..2..10 }这是一个有效的表达式:
this is a valid expression:
seqA.Where(even)但这不是:
seqA.All(even)为什么将 even 传递给 哪里允许但不允许All?
Why is passing even to Where allowed but not to All?
推荐答案虽然可能存在 bug,但我认为更好的方法是在处理 IEnumerable 在 F# 而不是 Linq 中
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更多推荐
将 F# 函数传递给 IEnumerable.Where 与 IEnumerable.All
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