我需要使用递归下降解析器构建表达式。
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,我有一个组合结构,每个终端都有一个类,并且我正在尝试按照上述语法的规则构建它。
我的想法是,我需要一个方法来处理每个语法规则,并且每个方法都必须构建其表达式树的一部分。
虽然使用更简单的语法(例如只允许布尔表达式的语法)可以工作,但我在这个语法中遇到了一些问题。
主要问题来自于 `expr` 规则,它迫使我甚至使用我的 + 和 - 的单元版本来允许诸如 `+3-4` 这样的表达式。这要求我找出何时将操作数视为一元运算符,何时视为二元运算符!
以下是我的 Builder 类的代码。 请注意,其中有些内容用意大利语编写,但我已经用注释解释了 EVERYTHING,包括我的问题。还要注意,有一行代码中我使用了伪代码,因此该行代码无法编译!
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
的计算结果为 a <= b && c > 1
a==[-4]
的计算结果为 a == a - 4
-4+3>c-2
的计算结果为 +3 > c - 2
可能还有其他错误,但这些是最常见的。所以我的问题是:
首先,您认为此代码是否存在逻辑问题?我的意思是,它是否按照语法执行,或者我做错了什么? 如何修复
expr
方法,使其能够处理一元和二元+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一个可行版本? 正如您在类名中看到的那样,这是我写的第六个实现,我真的不知道下一步该怎么做!
谢谢。