无法构造无限类型 - 实施尾递归阶乘计算器

Question

我想实现阶乘的尾递归版本：

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

我得到这个：

<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在这里？ （使用GHCI 7.0.1）

Answer 1

我不是一个强大的哈斯克尔程序员，但我想你想重写

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

为

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

由于(x-1, n * x)是一双类型，不是你想要的。

希望这会有所帮助！