有没有办法简化这条线？

Question

if reelID = reelWeights.Count - 1
    then Array.fold calc1 (0L,0) reelWeights.[reelID]
    else Array.fold calc2 (0L,0) reelWeights.[reelID]

我尝试使用管道,似乎放慢了一点(不知道为什么):

reelWeights.[reelID]
    |> (if reelID = reelWeights.Count - 1 then Array.fold calc1 else Array.fold calc2) (0L,0)

如果我做

let calc x = if x then calc1 else calc2
Array.fold (calc reelID = reelWeights.Count - 1) (0L,0) reelWeights.[reelID]

那么在循环中冗余检查条件的成本看起来很不错.

Answer 1

假设calc1并且calc2具有相同的签名(或者如果它们是值而不是函数,则是相同的类型):

let calc = if reelID = reelWeights.Count - 1 then calc1 else calc2
Array.fold calc (0L, 0) reelWeights.[reelID]