Prolog - 解析 [英] Prolog - Parsing

问题描述

我是语言序言的新手，并获得了有关在序言中解析的任务.我需要一些帮助来解决问题.

I'm new to the language prolog and have been given an assignment regarding parsing in prolog. I need some help in solving the problem.

在评估中，我们有语法:

In the assingment we have the grammar:

Expr ::= + Expr Expr | * Expr Expr | Num | Xer  
Xer  ::= x | ^ x Num  
Num  ::= 2 | 3 | .... a Integer (bigger than 1) ...

标记 ^ 与数学中的相同.5^5 等于 25.

The token ^ is the same as in math. 5^5 equals 25.

Parse 需要双向工作:使用实例化列表进行调用以生成 Ast，而具有实例化 Ast 的调用应该生成类似的前缀列表.

Parse needs to work both ways: a call with an instantiated list to generate an Ast, while a call with an instantiated Ast should generate similar prefix list.

我的评估说我需要做一个前缀解析来做到这一点:
示例(删除了 Ast 的值):

My assingment says that I need to make a prefix parse that does this:
Example(with the value of Ast removed):

?- parse([+, *, 2, x, ^, x, 5 ], Ast), parse(L, Ast).  
X = ...,  
L = [+, *, 2, x, ^, x, 5]  

我也想知道解析树的样子.

I would also like to know how the parse tree will look like.

推荐答案

Prolog 有一个特殊的形式来直接处理上下文无关文法:DCGs(定句文法).您的示例几乎立即转换为 DCG:

Prolog has a particular formalism to handle context-free grammars directly: DCGs (Definite Clause Grammars). Your example translates almost immediately into a DCG:

expr --> [+], expr, expr | [*], expr, expr | num | xer.

xer --> [x] | [^], [x], num.

num --> [2] | [3] | [4] | [5].

现在，您已经可以测试句子了:

Now, you already can test sentences:

?- phrase(expr, [+, *, 2, x, ^, x, 5 ]).
true ;
false.

?- phrase(expr, [+, *, *, 2, x, ^, x, 5 ]).
false.

您甚至可以像这样生成所有可能的句子:

You can even generate all possible sentences like so:

?- length(L, N), phrase(expr, L).
L = [2],
N = 1 ;
L = [3],
N = 1 ;
...

最后，您可以将抽象语法树添加到您的定义中.

And, finally, you can add the abstract syntax tree to your definition.

expr(plus(A,B)) --> [+], expr(A), expr(B).
expr(mul(A,B)) --> [*], expr(A), expr(B).
expr(Num) --> num(Num).
expr(Xer) --> xer(Xer).

xer(var(x)) --> [x].
xer(pow(var(x),N)) --> [^], [x], num(N).

num(num(2)) --> [2].
num(num(3)) --> [3].
num(num(4)) --> [4].
num(num(5)) --> [5].

所以现在你可以随意使用它了:

So now you can use it as desired:

?- phrase(expr(AST), [+, *, 2, x, ^, x, 5 ]), phrase(expr(AST),L).
AST = plus(mul(num(2), var(x)), pow(var(x), num(5))),
L = [+, *, 2, x, ^, x, 5] ;
false.

吹毛求疵:DCG 的接口谓词是 phrase/2 而不是 parse/2.

Just a nitpick: The interface predicate to DCGs is phrase/2 not parse/2.

这篇关于Prolog - 解析的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！