我正在尝试在ANTLR4中制作一个计算器，它可以使用几乎所有可能的符号作为数学运算符
具体：
-用户定义由运算符和优先级组成的操作。运算符可以是除某些系统符号(括号、逗号…)之外的任何符号组合。优先级是一个正整数。操作存储在java HashMap中
-有三种不同类型的操作：左侧(一元-减号，…)、右侧(阶乘，…)和二进制(加法，…)
-这些操作应该在运行时请求，以便在解析过程中(取消)激活操作。如果这不可能，那么应该在创建解析器时请求操作符
-对于优先级：最好是完全动态优先级(在运行时会请求遇到的操作的优先级)，但如果不可能，则应该有不同的优先级预设。(乘法、加法…)
我得到的是：
-用于运算符识别的工作代码
-优先级攀升代码，它生成了正确的解析树，但给出了一个错误：rule-expr失败的谓词：(getPriority($op)>=$_p)？
更新：修复了操作员识别代码，并找到了优先级攀升机制的代码
tokens { PREOP, POSTOP, BINOP, ERROR }
@lexer::members {
private static List<String> binaryOperators;
private static List<String> prefixOperators;
private static List<String> postfixOperators;
{
binaryOperators = new ArrayList<String>();
binaryOperators.add("+");
binaryOperators.add("*");
binaryOperators.add("-");
binaryOperators.add("/");
prefixOperators = new ArrayList<String>();
prefixOperators.add("-");
postfixOperators = new ArrayList<String>();
postfixOperators.add("!");
}
private Deque<Token> deque = new LinkedList<Token>();
private Token previousToken;
private Token nextToken;
@Override
public Token nextToken() {
if (!deque.isEmpty()) {
return previousToken = deque.pollFirst();
}
Token next = super.nextToken();
if (next.getType() != SYMBOL) {
return previousToken = next;
}
StringBuilder builder = new StringBuilder();
while (next.getType() == SYMBOL) {
builder.append(next.getText());
next = super.nextToken();
}
deque.addLast(nextToken = next);
List<Token> tokens = findOperatorCombination(builder.toString(), getOperatorType());
for (int i = tokens.size() - 1; i >= 0; i--) {
deque.addFirst(tokens.get(i));
}
return deque.pollFirst();
}
private static List<Token> findOperatorCombination(String sequence, OperatorType type) {
switch (type) {
case POSTFIX:
return getPostfixCombination(sequence);
case PREFIX:
return getPrefixCombination(sequence);
case BINARY:
return getBinaryCombination(sequence);
default:
break;
}
return null;
}
private static List<Token> getPrefixCombination(String sequence) {
if (isPrefixOperator(sequence)) {
List<Token> seq = new ArrayList<Token>(1);
seq.add(0, new CommonToken(MathParser.PREOP, sequence));
return seq;
}
if (sequence.length() <= 1) {
return null;
}
for (int i = 1; i < sequence.length(); i++) {
List<Token> seq1 = getPrefixCombination(sequence.substring(0, i));
List<Token> seq2 = getPrefixCombination(sequence.substring(i, sequence.length()));
if (seq1 != null & seq2 != null) {
seq1.addAll(seq2);
return seq1;
}
}
return null;
}
private static List<Token> getPostfixCombination(String sequence) {
if (isPostfixOperator(sequence)) {
List<Token> seq = new ArrayList<Token>(1);
seq.add(0, new CommonToken(MathParser.POSTOP, sequence));
return seq;
}
if (sequence.length() <= 1) {
return null;
}
for (int i = 1; i < sequence.length(); i++) {
List<Token> seq1 = getPostfixCombination(sequence.substring(0, i));
List<Token> seq2 = getPostfixCombination(sequence.substring(i, sequence.length()));
if (seq1 != null && seq2 != null) {
seq1.addAll(seq2);
return seq1;
}
}
return null;
}
private static List<Token> getBinaryCombination(String sequence) {
for (int i = 0; i < sequence.length(); i++) { // i is number of postfix spaces
for (int j = 0; j < sequence.length() - i; j++) { // j is number of prefix spaces
String seqPost = sequence.substring(0, i);
List<Token> post = getPostfixCombination(seqPost);
String seqPre = sequence.substring(sequence.length()-j, sequence.length());
List<Token> pre = getPrefixCombination(seqPre);
String seqBin = sequence.substring(i, sequence.length()-j);
if ((post != null || seqPost.isEmpty()) &&
(pre != null || seqPre.isEmpty()) &&
isBinaryOperator(seqBin)) {
List<Token> res = new ArrayList<Token>();
if (post != null)
res.addAll(post);
res.add(new CommonToken(MathParser.BINOP, seqBin));
if (pre != null)
res.addAll(pre);
return res;
}
}
}
return null;
}
/**
* Returns the expected operator type based on the previous and next token
*/
private OperatorType getOperatorType() {
if (isValueEnd(previousToken.getType())) {
if (isValueStart(nextToken.getType())) {
return OperatorType.BINARY;
}
return OperatorType.POSTFIX;
}
return OperatorType.PREFIX;
}
private enum OperatorType { BINARY, PREFIX, POSTFIX };
/**
* Checks whether the given token is a token found at the start of value elements
* @param tokenType
* @return
*/
private static boolean isValueStart(int tokenType) {
return tokenType == MathParser.INT;
}
/**
* Checks whether the given token is a token found at the end of value elements
* @param tokenType
* @return
*/
private static boolean isValueEnd(int tokenType) {
return tokenType == MathParser.INT;
}
private static boolean isBinaryOperator(String operator) {
return binaryOperators.contains(operator);
}
private static boolean isPrefixOperator(String operator) {
return prefixOperators.contains(operator);
}
private static boolean isPostfixOperator(String operator) {
return postfixOperators.contains(operator);
}
}
优先级攀升代码：
@parser::members {
static Map<String, Integer> precedenceMap = new HashMap<String, Integer>();
static {
precedenceMap.put("*", 2);
precedenceMap.put("+", 1);
precedenceMap.put("^", 4);
precedenceMap.put("-", 3);
precedenceMap.put("!", 5);
}
public static Integer getPrecedence(Token op) {
return precedenceMap.get(op.getText());
}
public static Integer getNextPrecedence(Token op) {
Integer p = getPrecedence(op);
if (op.getType() == PREOP) return p;
else if (op.getText().equals("^")) return p;
else if (op.getType() == BINOP) return p+1;
else if (op.getType() == POSTOP) return p+1;
throw new IllegalArgumentException(op.getText());
}
}
prog
: expr[0]
;
expr [int _p]
: aexpr
( {getPrecedence(_input.LT(1)) >= $_p}? op=BINOP expr[getNextPrecedence($op)]
| {getPrecedence(_input.LT(1)) >= $_p}? POSTOP
)*
;
atom
: INT
| '(' expr[0] ')'
| op=PREOP expr[getNextPrecedence($op)]
;
所以现在的问题是如何处理这个断言失败错误
感谢其他贡献者，我为我的问题找到了一个完整的(实际上相当干净的)解决方案。
运算符匹配：
通过查看遇到的一系列符号前后的标记，可以检测运算符的固定性。之后，应用一种算法来检测符号序列中的有效运算符序列。然后在令牌流中注入这些令牌(在nextToken()中)。只需确保在SYMBOL定义之前定义了所有硬编码标记即可。
优先级攀升：
事实上，这并没有那么难，这与ANTLR4的内部策略完全相同。
grammar Math;
tokens { PREOP, POSTOP, BINOP, ERROR }
@header {
import java.util.*;
}
@lexer::members {
private static List<String> binaryOperators;
private static List<String> prefixOperators;
private static List<String> postfixOperators;
{
binaryOperators = new ArrayList<String>();
binaryOperators.add("+");
binaryOperators.add("*");
binaryOperators.add("-");
binaryOperators.add("/");
System.out.println(binaryOperators);
prefixOperators = new ArrayList<String>();
prefixOperators.add("-");
System.out.println(prefixOperators);
postfixOperators = new ArrayList<String>();
postfixOperators.add("!");
System.out.println(postfixOperators);
}
private Deque<Token> deque = new LinkedList<Token>();
private Token previousToken;
private Token nextToken;
@Override
public Token nextToken() {
if (!deque.isEmpty()) {
return previousToken = deque.pollFirst();
}
Token next = super.nextToken();
if (next.getType() != SYMBOL) {
return previousToken = next;
}
StringBuilder builder = new StringBuilder();
while (next.getType() == SYMBOL) {
builder.append(next.getText());
next = super.nextToken();
}
deque.addLast(nextToken = next);
List<Token> tokens = findOperatorCombination(builder.toString(), getOperatorType());
for (int i = tokens.size() - 1; i >= 0; i--) {
deque.addFirst(tokens.get(i));
}
return deque.pollFirst();
}
private static List<Token> findOperatorCombination(String sequence, OperatorType type) {
switch (type) {
case POSTFIX:
return getPostfixCombination(sequence);
case PREFIX:
return getPrefixCombination(sequence);
case BINARY:
return getBinaryCombination(sequence);
default:
break;
}
return null;
}
private static List<Token> getPrefixCombination(String sequence) {
if (isPrefixOperator(sequence)) {
List<Token> seq = new ArrayList<Token>(1);
seq.add(0, new CommonToken(MathParser.PREOP, sequence));
return seq;
}
if (sequence.length() <= 1) {
return null;
}
for (int i = 1; i < sequence.length(); i++) {
List<Token> seq1 = getPrefixCombination(sequence.substring(0, i));
List<Token> seq2 = getPrefixCombination(sequence.substring(i, sequence.length()));
if (seq1 != null & seq2 != null) {
seq1.addAll(seq2);
return seq1;
}
}
return null;
}
private static List<Token> getPostfixCombination(String sequence) {
if (isPostfixOperator(sequence)) {
List<Token> seq = new ArrayList<Token>(1);
seq.add(0, new CommonToken(MathParser.POSTOP, sequence));
return seq;
}
if (sequence.length() <= 1) {
return null;
}
for (int i = 1; i < sequence.length(); i++) {
List<Token> seq1 = getPostfixCombination(sequence.substring(0, i));
List<Token> seq2 = getPostfixCombination(sequence.substring(i, sequence.length()));
if (seq1 != null && seq2 != null) {
seq1.addAll(seq2);
return seq1;
}
}
return null;
}
private static List<Token> getBinaryCombination(String sequence) {
for (int i = 0; i < sequence.length(); i++) { // i is number of postfix spaces
for (int j = 0; j < sequence.length() - i; j++) { // j is number of prefix spaces
String seqPost = sequence.substring(0, i);
List<Token> post = getPostfixCombination(seqPost);
String seqPre = sequence.substring(sequence.length()-j, sequence.length());
List<Token> pre = getPrefixCombination(seqPre);
String seqBin = sequence.substring(i, sequence.length()-j);
if ((post != null || seqPost.isEmpty()) &&
(pre != null || seqPre.isEmpty()) &&
isBinaryOperator(seqBin)) {
List<Token> res = new ArrayList<Token>();
if (post != null)
res.addAll(post);
res.add(new CommonToken(MathParser.BINOP, seqBin));
if (pre != null)
res.addAll(pre);
return res;
}
}
}
return null;
}
/**
* Returns the expected operator type based on the previous and next token
*/
private OperatorType getOperatorType() {
if (isAfterAtom()) {
if (isBeforeAtom()) {
return OperatorType.BINARY;
}
return OperatorType.POSTFIX;
}
return OperatorType.PREFIX;
}
private enum OperatorType { BINARY, PREFIX, POSTFIX };
/**
* Checks whether the current token is a token found at the start of atom elements
* @return
*/
private boolean isBeforeAtom() {
int tokenType = nextToken.getType();
return tokenType == MathParser.INT ||
tokenType == MathParser.PLEFT;
}
/**
* Checks whether the current token is a token found at the end of atom elements
* @return
*/
private boolean isAfterAtom() {
int tokenType = previousToken.getType();
return tokenType == MathParser.INT ||
tokenType == MathParser.PRIGHT;
}
private static boolean isBinaryOperator(String operator) {
return binaryOperators.contains(operator);
}
private static boolean isPrefixOperator(String operator) {
return prefixOperators.contains(operator);
}
private static boolean isPostfixOperator(String operator) {
return postfixOperators.contains(operator);
}
}
@parser::members {
static Map<String, Integer> precedenceMap = new HashMap<String, Integer>();
static {
precedenceMap.put("*", 2);
precedenceMap.put("+", 1);
precedenceMap.put("^", 4);
precedenceMap.put("-", 3);
precedenceMap.put("!", 5);
}
public static Integer getPrecedence(Token op) {
return precedenceMap.get(op.getText());
}
public static Integer getNextPrecedence(Token op) {
Integer p = getPrecedence(op);
if (op.getType() == PREOP) return p;
else if (op.getText().equals("^")) return p;
else if (op.getType() == BINOP) return p+1;
throw new IllegalArgumentException(op.getText());
}
}
prog
: expr[0]
;
expr [int _p]
: atom
( {getPrecedence(_input.LT(1)) >= $_p}? op=BINOP expr[getNextPrecedence($op)]
| {getPrecedence(_input.LT(1)) >= $_p}? POSTOP
)*
;
atom
: INT
| PLEFT expr[0] PRIGHT
| op=PREOP expr[getNextPrecedence($op)]
;
INT
: ( '0'..'9' )+
;
PLEFT : '(' ;
PRIGHT : ')' ;
WS
: [ trn]+ -> skip ; // skip spaces, tabs, newlines
SYMBOL
: .
;
注意：代码只是一个例子，而不是我真正的代码(运算符和优先级将从外部请求)
您不能在运行时为Antlr定义优先级/关联性规则。然而，您可以在解析中将所有运算符(语言内置或用户定义)解析为单链列表(如ArrayList<>)，然后在访问者中应用您自己的算法来获得优先级和关联性(如果您真的想的话，也可以在语法操作中)。
如果您多次迭代列表，那么算法本身并没有那么难。例如，您可以首先获取列表中每个运算符的优先级，然后检查优先级最高的运算符，查看其右侧或左侧是否关联，然后从中构建第一个(最底部)树节点。继续应用，直到列表为空，并且您已经构建了自己的"解析树"，但没有解析(您不再使用抽象输入字符串)。
或者，在运行时对Antlr进行外部调用以编译.g4，并对javac进行外部调用来编译生成的Antlr代码，然后使用反射来调用它。然而，这可能要慢得多，而且可能更难实现。
根据Symbol优先级的某些运行时定义，可以"正确"工作的解析器规则。虽然最初看起来不是一个惯用的选择，但将语义分析从解析器中延迟出来的标准替代方案会产生一个非常差的解析树——这使其成为标准设计规则的合理例外。
在(过于简化的)形式中，解析器规则是：
expr : LParen expr RParen # group
| expr Symbol expr # binary
| expr Symbol # postfix
| Symbol expr # prefix
| Int+ # value
;
为了解决歧义，添加内联谓词：
expr : LParen expr RParen # group
| expr s=Symbol { binary($s) }? expr # binary
| expr s=Symbol { postfix($s) }? # postfix
| s=Symbol { prefix($s) }? expr # prefix
| Int+ # value
;
对于任何给定的Symbol，单个谓词方法的计算结果应为true。
扩展到多个Symbol字符串会增加一点复杂性(例如，区分二进制和后跟前缀的后缀)，但机制基本保持不变。
我认为您的方法是正确的。我建议使用以下语法：
grammar Op;
options {
superClass=PrecedenceParser;
}
prog : expr[0] ;
expr[int _p] locals[Token op]: INT ({$op = _input.LT(1);} {getPrecedence($op) >= $_p}? OP expr[getPrecedence($op)])*;
INT : ( '0'..'9' )+ ;
OP : '+' | '*'; // all allowed symbols, should be extended
WS : [ trn]+ -> skip ; // skip spaces, tabs, newlines
op的规则应包含所有允许的运算符符号。我对+和*的限制只是为了简单起见。解析器的超级类是：
public abstract class PrecedenceParser extends Parser {
private Map<String, Integer> precedences;
public PrecedenceParser(TokenStream input) {
super(input);
this.precedences = new HashMap<>();
}
public PrecedenceParser putOperator(String op, int p) {
precedences.put(op, p);
return this;
}
public int getPrecedence(Token operator) {
Integer p = precedences.get(operator.getText());
if (p == null) {
return Integer.MAX_VALUE;
} else {
return p;
}
}
}
结果
有{+ : 4, * : 3 }先例
(prog (expr 1 + (expr 2) * (expr 3 + (expr 4))))
有{+ : 3, * : 4 }先例
(prog (expr 1 + (expr 2 * (expr 3) + (expr 4))))
从左到右评估这些序列相当于按优先级评估它们。
这种方法应该适用于较大的运算符集。ANTLR4在内部使用这种方法进行优先级攀升，但ANTLR使用常量而不是优先级映射(因为它假设优先级是在解析器构建时确定的)。