@parser-generator/core/lib/esm/lr/lr-parser.js

A Parser Generator that supports LL,SLR,LR1,LALR

import { MismatchError } from "../utils";
import { Stack } from "@light0x00/shim";
import _debug from "debug";
let debug = _debug("PG:lr-parser");
export class LRParser {
    constructor(parsingTable) {
        this.parsingTable = parsingTable;
    }
    parse(lexer) {
        let stateStack = new Stack();
        stateStack.push(0);
        let astStack = new Stack();
        let lookahead = lexer.peek();
        while (1) {
            let topSta = stateStack.peek();
            let op = this.parsingTable.get(topSta, lookahead.key());
            if (op == undefined)
                throw new MismatchError(this.parsingTable.getExpectedTokens(topSta), lookahead);
            if (op.isShift()) {
                astStack.push(lexer.next());
                stateStack.push(op.nextStateId);
                lookahead = lexer.peek();
            }
            else if (op.isGoto()) {
                stateStack.push(op.nextStateId);
                lookahead = lexer.peek();
            }
            else if (op.isReduce()) {
                /* 此次归约产生的AST节点所需的元素 */
                let eles = [];
                // 动作: 归约完成后将符号对应的状态弹出
                // 每个状态都由输入一个符号得到 因此每个状态都一个对应的符号  详见:P158
                for (let i = 0; i < op.prod.body.length; i++) {
                    stateStack.pop();
                    eles.unshift(astStack.pop()); //issue AST元素的顺序问题
                }
                astStack.push(op.prod.postAction(eles, astStack)); //issue ASTNode的元素的顺序问题
                debug(`reduce ${op.prod}, make ast: ${astStack.peek()}`);
                lookahead = op.prod.head;
            }
            else if (op.isAccept()) {
                debug(`accept!`);
                break;
            }
            debug(`${astStack.join(" ")}`);
        }
        return astStack.pop();
    }
}
/*
1. ASTNode的元素的顺序问题

考虑如下文法
    S->E
    E->E+T | E-T | T
    T->T*F | T/F |F
    F->(E) | NUM

假设输入:
1 + 2 / 3
必然在某一时刻 Ast Stack中的元素为:
ENode + TNode / FNode

此时,即将要执行的动作是将 TNode / FNode 归约为T,创建一个TNode:
    new TNode([TNode,'/',FNode])
这时要注意的是,传给TNode的集合应该保持与产生式中定义的符号顺序保持一致.
假如传入的是 [FNode,'/',TNode], 那么原本的2/3的含义将变为3/2
*/
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