haskell - 使用列表理解生成元组

Question

我使用列表理解生成元组:

[(a,b,c) | a <- [1..3], b <- [1..3], c <- [1..3]]

由于元组的所有三个条目都来自同一区间，是否有更短的写法？

Answer 1

一些选择

let xs = [1..3] in [(a,b,c) | a <- xs, b <- xs, c <- xs]
[(a,b,c) | let xs = [1..3], a <- xs, b <- xs, c <- xs]
(,,) <$> [1..3] <*> [1..3] <*> [1..3]
let xs = [1..3] in (,,) <$> xs <*> xs <*> xs
(\[a,b,c]->(a,b,c)) <$> replicateM 3 [1..3]

但是，我会寻找最易读的，而不是最短的。