Question

我正在尝试实现factorial的尾递归版本：

let{factorial 0 n = n; factorial x n =  factorial (x-1, n * x)}

我明白了：

<interactive>:1:41:
Occurs check: cannot construct the infinite type: t1 = t1 -> t1
In the return type of a call of `factorial'
In the expression: factorial (x - 1, n * x)
In an equation for `factorial':
    factorial x n = factorial (x - 1, n * x)

<interactive>:1:52:
Occurs check: cannot construct the infinite type: t0 = (t0, t1)
In the first argument of `(-)', namely `x'
In the expression: x - 1
In the first argument of `factorial', namely `(x - 1, n * x)'

<interactive>:1:61:
Occurs check: cannot construct the infinite type: t1 = (t0, t1)
In the second argument of `(*)', namely `x'
In the expression: n * x
In the first argument of `factorial', namely `(x - 1, n * x)'

我如何构建无限类型？（使用GHCi 7.0.1）

Answer 1

我不是一个强大的Haskell程序员，但我认为你想重写

factorial x n =  factorial (x-1, n * x)

作为

factorial x n =  factorial (x-1) (n * x)

由于(x-1, n * x)是一种配对类型，这不是您想要的。

希望这有帮助！

无法构造无限类型 - 实现尾递归因子计算器

1 个答案: