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>→<factb>[AND <factb>]*
<factb>→<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 “&&”
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的规则来构建它。
我的想法是每个语法规则需要一个方法，每个方法都必须构建表达式树的一部分。
虽然它使用更简单的语法（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expr规则，它允许我使用我的+和 - 的一元版本来工作，方法是允许+3-4这样的表达式。这需要我找出何时必须将操作数视为一元和二进制！
这是我的Builder类的代码。 请注意，有些东西是意大利语，但我已经使用评论解释了所有问题，甚至是我的问题。另请注意，有一行我使用了伪代码，所以这个不可编辑！
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
可能还有一些错误，但这些错误最常见。
所以这是我的问题：
首先，您认为此代码存在一些逻辑问题吗？我的意思是，它是做了语法所说的，还是我做了一些非常错误的事情？
您如何修复expr方法以使其适用于一元和二元+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那样，这是我写的第六个实现，我真的不知道接下来该做什么！
感谢。
答案 0 :(得分：5)
我认为你在实现递归下降解析器时遇到问题，因为你的语法非常复杂。由[ ... ]*形式表示的任意迭代很难绕过你的脑袋。试着这样思考：
<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 → ""
此语法与您发布的语法相同，但我将[ ... ]*规则拆分为自己的非终端。为此，我添加了EPSILON规则，允许非终端匹配""（空字符串）。这与您的[ ... ]规则基本相同，可能会或可能不会与某些内容匹配。 epsilon规则只是最后一种选择，即“现在停止尝试匹配”。例如，如果您当前正在解析cond-tail，那么您首先要检查输入上的OR。如果您没有看到OR，那么您将转到下一个替代方案EPSILON。这意味着此条件链中不再有OR个条件，因此cond-tail匹配空字符串。
您可以通过在expr-tail和term-tail中将运算符组合在一起来进一步简化语法：
<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，但至少应使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正在执行哪些操作。）
现在应该非常明显+和-在哪些情况下是一元运算符，因为它们将由您的新buildUnaryOp方法解析（对应于新的unary-op语法中的规则）。
对于新的*-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烧掉堆栈，则可以在函数体内使用循环实现它们。 / p>
这是你的主要问题：
/**
* 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。我从来没有见过一个递归下降的解析器，其方法的返回类型为void - 至少在它实际构建一个抽象语法树时没有。 example on wikipedia有void个返回类型，但这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，因此在每个同级调用buildExpr()中都会覆盖该字段，并且您正在丢失数据。无论如何，那是我的理论。您可以通过制作expr长"1+2+3+4+5"字符串来测试它。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。如果您对此更清楚，您仍然可以使用变异。这是重构的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;
}
以下是新build()的样子：
public Espressione build() {
return buildCond();
}
如果你重新编写其余的代码来以这种方式构建树（而不是使用实例级字段来保存你的状态）那么它应该工作得更好。
祝你好运！