数据类型问题中的标准 ml 函数

发布于 2024-08-20 05:15:08 字数 489 浏览 8 评论 0原文

我必须创建一个有关 peano 数的函数，定义为以下数据类型：

datatype 'a peano = P of ('a -> 'a) * 'a -> 'a
val zero = P(fn (f, x) => x)

我必须实现的函数查找 peano 参数 P(p) 的连续 peano 数。这就是我所写的：

fun suc (P(p)) = case P(p) of P(fn(f,x)=>x) => P(fn(f,x)=>f(x));

问题是我收到这些错误：

stdIn:4.33-4.36 Error: syntax error: deleting FN LPAREN
stdIn:4.43 Error: syntax error found at RPAREN

我不知道我做错了什么。请帮忙！

I have to create a function about peano numbers defined as the following datatype:

datatype 'a peano = P of ('a -> 'a) * 'a -> 'a
val zero = P(fn (f, x) => x)

The function that I have to implement finds the succesive peano number of the peano parameter P(p). This is what I have written:

fun suc (P(p)) = case P(p) of P(fn(f,x)=>x) => P(fn(f,x)=>f(x));

The problem is that i get these errors:

stdIn:4.33-4.36 Error: syntax error: deleting FN LPAREN
stdIn:4.43 Error: syntax error found at RPAREN

I don't know what Im doing wrong. Please help!

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仅此而已 2024-08-27 05:15:08

这段代码有很多问题。编译器抱怨的问题是，您

fn (f,x) => x

在 case 臂的左侧有一个函数定义，其中只允许使用模式。

其他一些问题：

多余的括号使代码难以阅读（可以提供删除它们的建议）。
您的 case 表达式是多余的；在函数定义中
```
 fun suc (P p) = ...
```
应该可以仅使用 p 进行计算，而无需进行任何案例分析。
由于 P 带有一个函数，如果你这样写，可能会更容易
```
 fun suc (P f) = ...
```
并确保在结果中，f 应用于一对（根据数据类型声明的要求）。

There are a number of problems in this code. The one the compiler is whining about is that you have a function definition

fn (f,x) => x

on the left-hand side of a case arm, where only patterns are permitted.

Some other problems:

Redundant parentheses make the code hard to read (advice is available on removing them).
Your case expression is redundant; in the function definition
```
 fun suc (P p) = ...
```
it should be possible just to compute with p without any more case analysis.
Since P carries a function, you will probably have an easier time if you write
```
 fun suc (P f) = ...
```
and make sure that in the result, f is applied to a pair (as required by the datatype declarations).