如何将逻辑表示法转换为Haskell语法 [英] How to translate logical notation to Haskell syntax

问题描述

我最近在uni上找到了Haskell，我正在通过一系列练习来工作，下面是我无法理解的一个片段：

I've recently picked up Haskell at uni and I'm working my way through a set of exercises, here's a snippet of one that I can't make sense of:

考虑以下简单的前缀计算器语言的语法：

"Consider the following grammar for a simple, prefix calculator language:"

num ::= 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
int ::= num | num int
expr ::= int | - expr | + expr expr | * expr expr

我很困惑如何将它转换成Haskell语法（我是一个完整的新手在Haskell和函数式编程，请温柔）

I'm confused as how to translate this into Haskell syntax (I'm a complete beginner in both Haskell and functional programming, please be gentle)

我怀疑 num ， int 和 expr 都是假设可以使用 data 声明的类型/值>或键入，并且它们对计算器施加约束。然而，我无法理解：我如何声明类型或数据（不是变量）值，即0-9？另外，如何在声明中放置 - 或 + 等符号？

I suspect that num, int and expr are all, supposedly, types/values that can be declared using data or type and that they impose constraints on the calculator. However I can't make sense of either: How do I declare type or data(not a variable) for fixed values, namely 0-9? Also, how can I put symbols like - or + in a declaration?

推荐答案

不要混淆表示AST的语法中的字符串。比较字符串

Don't confuse a string in the grammar for the AST that represents it. Compare the string

"+ + 3 4 5"

这是你给出的文法中的一个字符串

which is a string in the grammar you've been given with

Plus (Plus (Literal 3) (Literal 4)) (Literal 5)

这是一个合理的可以解析为 String 的AST的Haskell值。

which would be a sensible Haskell value for the AST that String could get parsed to.

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