我知道我们可以在 Haskell 中使用以下模式匹配:
```haskell ```
```haskell ```
sum :: (Num a) => [a] -> a
sum [] = 0
sum (x:xs) = x + sum xs
But why can’t we use [x] ++ xs
?
sum :: (Num a) => [a] -> a
sum [] = 0
sum ([x] ++ xs) = x + sum xs
:
是一个构造函数吗? - 4castle1:2:3:4:[]
的简写形式。 - Julia Path(\(x++y) -> y++x) "abcde"
会得出什么结果? - chi