java - 为我的语法编写递归后代解析器

Question

我必须使用 Recursive Descendent Parser Builder 构建表达式。

这是我的语法：

----RULES----
<cond> → <termb> [OR <termb>]*
<termb>→&lt;factb>[AND <factb>]*
<factb>→&lt;expr> RELOP <expr> | NOT <factb> | OPAR <cond> CPAR
<expr> → [PLUS | MINUS] <term> [(PLUS <term>) | (MINUS <term>)]*
<term> → <termp> [(MULT <termp>) | (DIV <termp>) | (REM <termp>)]*
<termp> → <fact> [POWER <fact>]*
<fact> → ID | NUM | OPAR1 <expr> CPAR1
----TERMINALS----
ID → ("A" | ... | "Z" | "a" | ...| "z") [("A"| ... | "Z" | "a" | ...| "z" | "0" | ... | "9")]*
NUM → ("0" | ... | "9") [("0" | ... | "9")]*
OPAR → "("
CPAR → ")"
OPAR1 → "["
CPAR1 → "]"
RELOP → EQ | NEQ | GT | GE | LT | LE
EQ → "= ="
NEQ → "!="
GT → ">"
GE → ">="
LT → "<"
LE → "<="
POWER → "^"
DIV → "/"
REM → "%"
MULT → "*"
MINUS → "−"
PLUS → "+"
AND → “and” or “&amp;&”
OR → “or” or “||”
NOT → “not” or “!”

现在，我有一个复合结构，每个终端都有一个类，我正在尝试按照上述语法规则来构建它。

这个想法是我需要一个用于每个语法规则的方法，并且每个方法都必须构建表达式树的一部分。

虽然它适用于更简单的语法（例如，只允许布尔表达式的语法），但我遇到了一些问题。

主要问题来自于expr规则，它强制我使用 + 和 - 的一元版本，允许这样的表达式+3-4。这需要我找出何时必须将操作数视为一元以及何时视为二进制！

这是我的 Builder 类的代码。请注意，有些东西是意大利语的，但我已经使用评论解释了一切，甚至是我的问题。另请注意，有一行我使用了伪代码，所以这一行不可编译！

package espressioniCondizionali;

import espressioniCondizionali.espressioni.*;

/**
 *
 * @author Stefano
 */
public class Builder6 {

    /**
     * This one is used just to parse the input String.
     * It has a prossimoSimbolo() method (something like a nextSymbol()) that returns a String with the next Symbol in the String
     */
    private GestoreInput gestoreInput;
    /**
     * Espressione is the Composite structure that represents and expression
     */
    private Espressione root = null;
    private String symbol = null;

    public Builder6(GestoreInput gestoreInput) {
        this.gestoreInput = gestoreInput;
    }

    public Espressione build() {
        buildCond();
        return root;
    }

    private void buildCond() {
        buildTermb();
        //I'm unsing a class named Simboli that holds every terminal symbol of my grammar. Symbol names are the same defined in the grammar.
        while (symbol.equalsIgnoreCase(Simboli.OR1) || symbol.equalsIgnoreCase(Simboli.OR2)) {
            Or or = new Or();
            or.setLeft(root);
            buildTermb();
            or.setRight(root);
            root = or;
        }
    }

    private void buildTermb() {
        buildFactb();
        while (symbol.equalsIgnoreCase(Simboli.AND1) || symbol.equalsIgnoreCase(Simboli.AND2)) {
            And and = new And();
            and.setLeft(root);
            buildFactb();
            and.setRight(root);
            root = and;
        }
    }

    private void buildFactb() {
        buildExpr();
        if (symbol.equalsIgnoreCase(Simboli.EQ) || symbol.equalsIgnoreCase(Simboli.NEQ) || symbol.equalsIgnoreCase(Simboli.GT) || symbol.equalsIgnoreCase(Simboli.LT) || symbol.equalsIgnoreCase(Simboli.LE) || symbol.equalsIgnoreCase(Simboli.GE)) {
            Operatore op = null;
            switch (symbol) {
                case Simboli.EQ: {
                    op = new Eq();
                    break;
                }
                case Simboli.NEQ: {
                    op = new Neq();
                    break;
                }
                case Simboli.GT: {
                    op = new Gt();
                    break;
                }
                case Simboli.GE: {
                    op = new Ge();
                    break;
                }
                case Simboli.LT: {
                    op = new Lt();
                    break;
                }
                case Simboli.LE: {
                    op = new Le();
                    break;
                }
                default: {
                    //Symbol not recognized, abort!
                    throw new RuntimeException("\"" + symbol + "\" non è un simbolo valido.");
                }
            }
            op.setLeft(root);
            buildExpr();
            op.setRight(root);
            root = op;
        } else if (symbol.equalsIgnoreCase(Simboli.NOT1) || symbol.equals(Simboli.NOT2)) {
            Not not = new Not();
            buildFactb();
            not.setRight(root);
            root = not;
        } else if (symbol.equalsIgnoreCase(Simboli.OPAR)) {
            buildCond();
            if (!symbol.equalsIgnoreCase(Simboli.CPAR)) { //If there's no closgin square bracket it means that our expression is not well formed
                throw new RuntimeException("Espressione non valida. Attesa una \")\", trovato \"" + symbol + "\"");
            }
        }
    }

    private void buildExpr() {
        Operatore op = null;
        if (symbol != null) { //Let's check if our expression starts with a + or a -
            if (symbol.equalsIgnoreCase(Simboli.PLUS)) {
                op = new Plus();
            } else if (symbol.equalsIgnoreCase(Simboli.MINUS)) {
                op = new Minus();
            }
        }
        buildTerm();
        //If our op is still null, we didn't found a + or a - so our operand will be a binary one
        //If op != null, our + or - is unary and we've got to manage it! A unary operand doesn't have a left son!
        if (op != null) {
            op.setRight(root);
            root = op;
        }
        //Since we can have something like -3+2+s-x we've got to check it
        while (symbol.equalsIgnoreCase(Simboli.PLUS) || symbol.equalsIgnoreCase(Simboli.MINUS)) {
            Operatore op1 = null; //We define a new Operatore that will be a + or a -
            switch (symbol) {
                case Simboli.PLUS: {
                    op1 = new Plus();
                    break;
                }
                case Simboli.MINUS: {
                    op1 = new Minus();
                    break;
                }
            }
            /*
             * Here comes a BIG problem. We used the first if/else to check if
             * our unary operand was at the beginning of the string, but now
             * we've got to see if our current operand is either binary or
             * unary! That's because we can have both a==-1+d and a==3-1+d. In
             * the first case, the - is unary, while is binary in the second.
             * So, how do we choose it?
             * 
             * EXAMPLE : (-a>2 || -b-12>0)
             * This one will be evaluated to -a > 2 || -12 > 0 that's clearly wrong!
             * -b is missing before -12. That's because the -12 is used as unary 
             * and so it won't have a left child (the -b part)
             */
            //--PSEUDOCODE
            if (op1 is not unary) {
                op1.setLeft(root);
            }
            //--END PSEUDOCODE
            //CURRENT IMPLEMENTATION OF THE PSEUDOCODE PART
            if (root != null && (root.getClass().equals(Num.class) || root.getClass().equals(Id.class))) { //It is binary if the previous value is a constant or a variable but not if it is an operand!                  
                op1.setLeft(root);
            }
            //END OF CURRENT IMPLEMENTATION OF THE PSEUDOCODE PART
            //Setting the right child must be done in both cases
            buildTerm();
            op1.setRight(root);
            root = op1;
        }
    }

    private void buildTerm() {
        buildTermp();
        while (symbol.equalsIgnoreCase(Simboli.MULT) || symbol.equalsIgnoreCase(Simboli.DIV) || symbol.equalsIgnoreCase(Simboli.REM)) {
            Operatore op = null;
            switch (symbol) {
                case Simboli.MULT: {
                    op = new Mult();
                    break;
                }
                case Simboli.DIV: {
                    op = new Div();
                    break;
                }
                case Simboli.REM: {
                    op = new Rem();
                    break;
                }
            }
            op.setLeft(root);
            buildTermp();
            op.setRight(root);
            root = op;
        }
    }

    private void buildTermp() {
        buildFact();
        while (symbol.equalsIgnoreCase(Simboli.POWER)) {
            Power p = new Power();
            p.setLeft(root);
            buildFact();
            p.setRight(root);
            root = p;
        }
    }

    private void buildFact() {
        symbol = gestoreInput.prossimoSimbolo();
        if (symbol.equalsIgnoreCase(Simboli.OPAR1)) { //Sottoespressione            
            buildExpr();
            if (!symbol.equalsIgnoreCase(Simboli.CPAR1)) { //Se non c'è una parentesi chiusa vuol dire che l'espressione non valida
                throw new RuntimeException("Espressione non valida. Attesa una \"]\", trovato \"" + symbol + "\"");
            }
        } else if (symbol.matches("[A-Za-z][A-Za-z | 0-9]*")) { //Nome di una variabile!
            root = new Id(symbol);
            symbol = gestoreInput.prossimoSimbolo();
        } else if (symbol.matches("[0-9]*")) { //E' una costante!
            root = new Num(symbol);
            symbol = gestoreInput.prossimoSimbolo();
        }
    }
}

已知问题：

(a<=b && c>1) || a==4评估为a <= b && c > 1

a==[-4]评估为a == a - 4

-4+3>c-2评估为+3 > c - 2

可能还有更多错误，但这些是最常见的。

所以这是我的问题：

首先，你认为这段代码有一些逻辑问题吗？我的意思是，它是按照语法所说的还是我做错了什么？您将如何修复该expr方法以使其适用于一元和二进制 + 或 - 操作数？

如果我的代码完全错误，是否有人可以帮助我编写工作版本？正如你在类名中看到的，这是我写的第六个实现，我真的不知道接下来要做什么！

谢谢。

Answer 1

第 1 步：以不同的方式思考语法

我认为您在实现递归下降解析器时遇到问题，因为您的语法非常复杂。表单所代表的任意迭代[ ... ]*很难让你理解。试着这样想：

<cond> → <termb> <cond-tail>
<cond-tail> → OR <termb> <cond-tail> | EPSILON
<termb> → <factb> <termb-tail>
<termb-tail> → AND <factb> <termb-tail> | EPSILON
<factb> → <expr> RELOP <expr> | NOT <factb> | OPAR <cond> CPAR
<expr> → <unary-op> <term> <expr-tail>
<unary-op> → PLUS | MINUS | EPSILON
<expr-tail> → ((PLUS <term>) | (MINUS <term>)) <expr-tail> | EPSILON
<term> → <termp> <term-tail>
<term-tail> → ((MULT <termp>) | (DIV <termp>) | (REM <termp>)) <term-tail> | EPSILON
<termp> → <fact> <termp-tail>
<termp-tail> → POWER <fact> <termp-tail> | EPSILON
<fact> → ID | NUM | OPAR1 <expr> CPAR1
EPSILON → ""

该语法与您发布的语法相同，但我将[ ... ]*规则拆分为它们自己的非终端。为了做到这一点，我添加了EPSILON规则，它允许非终端匹配""（空字符串）。这与您的规则基本相同[ ... ]，可能实际匹配也可能不匹配。epsilon 规则只是表示“现在停止尝试匹配”的最后一个替代方案。例如，如果您当前正在解析 acond-tail那么您将首先检查OR输入中的 an。如果您没有看到 anOR那么您将继续使用下一个替代方案，即EPSILON. OR这意味着这个条件链中没有更多的条件，所以它cond-tail匹配空字符串。

expr-tail您可以通过在and中将运算符组合在一起来进一步简化语法term-tail：

<cond> → <termb> <cond-tail>
<cond-tail> → OR <termb> <cond-tail> | EPSILON
<termb> → <factb> <termb-tail>
<termb-tail> → AND <factb> <termb-tail> | EPSILON
<factb> → <expr> RELOP <expr> | NOT <factb> | OPAR <cond> CPAR
<expr> → <unary-op> <term> <expr-tail>
<unary-op> → PLUS | MINUS | EPSILON
<expr-tail> → (PLUS | MINUS) <term> <expr-tail> | EPSILON
<term> → <termp> <term-tail>
<term-tail> → (MULT | DIV | REM) <termp> <term-tail> | EPSILON
<termp> → <fact> <termp-tail>
<termp-tail> → POWER <fact> <termp-tail> | EPSILON
<fact> → ID | NUM | OPAR1 <expr> CPAR1
EPSILON → ""

现在您应该返回并重构您的解决方案，为每个新规则添加新方法。（您可以选择将新规则的代码保留在原始规则中，但至少用注释清楚地标记不同的部分，以帮助您跟踪代码的哪些部分在做什么。）

现在应该很明显在哪些情况下+和-是一元运算符，因为它们将被您的新buildUnaryOp方法解析（对应unary-op于语法中的新规则）。

至于新*-tail规则，您可以选择严格递归地实现它们，或者如果您担心在长表达式上炸毁堆栈，您可以在函数体内使用循环来实现它们。

第 2 步：修复您的状态跟踪

这是您的主要问题：

/**
 * Espressione is the Composite structure that represents and expression
 */
private Espressione root = null;
private String symbol = null;

您正在使用实例级字段来存储递归下降解析器的状态。我从未见过一个递归下降解析器，其方法的返回类型为void——至少在它实际上构建抽象语法树时不会。wikipedia上的示例具有void返回类型，但那是因为它所做的只是检查输入并丢弃结果，而不是实际从输入构建 AST。

由于您将状态存储在类级字段中，因此该字段在每个同级调用中都会被覆盖，buildExpr()并且您正在丢失数据。无论如何，这是我的理论。您可以通过制作一长串exprs like来测试它"1+2+3+4+5"。它应该最终放弃前面的所有条款。

我建议在功能上构建解析器（不使用任何.set*()方法），其中每个build*()方法都从 AST 返回解析的节点。这将涉及更改所有构造函数以接受子节点。如果您更清楚，您仍然可以使用突变代替。这是一个重构buildCond：

private Espressione buildCond() {
      Espressione leftHandSide = buildTermb();
      if (symbol.equalsIgnoreCase(Simboli.OR1) || symbol.equalsIgnoreCase(Simboli.OR2)) {
          Or or = new Or();
          or.setLeft(leftHandSide);
          or.setRight(buildTermb());
          return or;
      }
      return leftHandSide;
  }

这是新的build()样子：

public Espressione build() {
    return buildCond();
}

如果您以这种方式重新编写其余代码以构建树（而不是使用实例级字段来保存您的状态），那么它应该会更好地工作。

祝你好运！