java – 为我的语法编写一个递归后代解析器
作者：互联网
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,我正在尝试按照上面的语法规则来构建它.
我的想法是每个语法规则都需要一个方法,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,它迫使我即使使用我的一元版本也可以工作 – 通过允许表达式如3-4.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< =&& c> 1)|| a == 4评估为< = 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？
你如何修复expr方法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？
正如您在类名中看到的,这是我编写的第六个实现,我真的不知道下一步该做什么！
谢谢.
解决方法:
第1步：以不同的方式思考语法
我认为你在实现你的递归下降解析器时遇到了问题,因为你的语法非常复杂.由[…] *形式表示的任意迭代很难包裹.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 → ""
这个语法与你发布的语法相同,但我把[…] *规则分成了自己的非终端.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).这基本上与您的规则相同,可能会或可能不会实际匹配某些内容. 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.如果你没有看到OR,那么你将转向下一个替代品,即EPSILON.这意味着此条件链中不再有OR条件,因此这个cond-tail在空字符串上匹配.
您可以通过将运算符组合在expr-tail和term-tail中来进一步简化语法：
<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.)
现在它应该是非常明显的在哪些情况和 – 是一元运算符,因为它们将被你的新buildUnaryOp方法解析(对应于语法中的新的unary-op规则).
至于新的* -tail规则,你可以选择严格递归地实现它们,或者如果你担心在长表达式上烧掉你的堆栈,你可以在函数体内用循环来实现它们.
第2步：修复状态跟踪
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 – 至少在它实际构建抽象语法树时是这样. example on wikipedia具有void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,因此在每个兄弟调用buildExpr()时都会覆盖该字段,并且您将丢失数据.无论如何,那是我的理论.您可以通过制作一长串exprs来测试它,例如“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;
}
这是new build()的样子：
public Espressione build() {
return buildCond();
}
如果您重新使用其余代码以这种方式构建树(而不是使用实例级字段来保存您的状态),那么它应该工作得更好.
祝好运！
标签：recursive-descent,builder,java,parsing 来源： https://codeday.me/bug/20190901/1786363.html