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可能使用正则表达式验证中缀表示法

发布于 2024-11-03 08:25:13 字数 133 浏览 6 评论 0原文

我正在考虑验证由字母作为操作数和 +-*/$ 作为运算符组成的中缀表示法 [例如：A+B-(C/D)$(E+F)] 在 Java 中使用正则表达式。还有更好的办法吗？有我可以使用的正则表达式模式吗？

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智商已欠费 2024-11-10 08:25:14

我不熟悉 infix 的语言语法，但是您当然可以进行首次通过验证检查，它只是验证字符串中的所有字符是否有效（即可接受的字符 = AZ、+、-、*、/、$、 （ 和 ））。这是一个检查有效字符的 Java 程序，还包含一个检查不平衡（可能嵌套）括号的函数：

import java.util.regex.*;
public class TEST {
    public static void main(String[] args) {
        String s = "A+B-(C/D)$(E+F)";
        Pattern regex = Pattern.compile(
            "# Verify that a string contains only specified characters.\n" +
            "^                # Anchor to start of string\n" +
            "[A-Z+\\-*/$()]+  # Match one or more valid characters\n" +
            "$                # Anchor to end of string\n",
            Pattern.COMMENTS);
        Matcher m = regex.matcher(s);
        if (m.find()) {
            System.out.print("OK: String has only valid characters.\n");
        } else {
            System.out.print("ERROR: String has invalid characters.\n");
        }
        // Verify the string contains only balanced parentheses.
        if (checkParens(s)) {
            System.out.print("OK: String has no unbalanced parentheses.\n");
        } else {
            System.out.print("ERROR: String has unbalanced parentheses.\n");
        }
    }
    // Function checks is string contains any unbalanced parentheses.
    public static Boolean checkParens(String s) {
        Pattern regex = Pattern.compile("\\(([^()]*)\\)");
        Matcher m = regex.matcher(s);
        // Loop removes matching nested parentheses from inside out.
        while (m.find()) {
            s = m.replaceFirst(m.group(1));
            m.reset(s);
        }
        regex = Pattern.compile("[()]");
        m = regex.matcher(s);
        // Check if there are any erroneous parentheses left over.
        if (m.find()) {
            return false;   // String has unbalanced parens.
        }
        return true;        // String has balanced parens.
    }
}

这不会验证语法，但作为过滤掉明显错误字符串的第一个测试可能很有用。

I am not familiar with the language syntax of infix, but you can certainly do a first pass validation check which simply verifies that all of the characters in the string are valid (i.e. acceptable characters = A-Z, +, -, *, /, $, ( and )). Here is a Java program which checks for valid characters and also includes a function which checks for unbalanced (possibly nested) parentheses:

import java.util.regex.*;
public class TEST {
    public static void main(String[] args) {
        String s = "A+B-(C/D)$(E+F)";
        Pattern regex = Pattern.compile(
            "# Verify that a string contains only specified characters.\n" +
            "^                # Anchor to start of string\n" +
            "[A-Z+\\-*/$()]+  # Match one or more valid characters\n" +
            "$                # Anchor to end of string\n",
            Pattern.COMMENTS);
        Matcher m = regex.matcher(s);
        if (m.find()) {
            System.out.print("OK: String has only valid characters.\n");
        } else {
            System.out.print("ERROR: String has invalid characters.\n");
        }
        // Verify the string contains only balanced parentheses.
        if (checkParens(s)) {
            System.out.print("OK: String has no unbalanced parentheses.\n");
        } else {
            System.out.print("ERROR: String has unbalanced parentheses.\n");
        }
    }
    // Function checks is string contains any unbalanced parentheses.
    public static Boolean checkParens(String s) {
        Pattern regex = Pattern.compile("\\(([^()]*)\\)");
        Matcher m = regex.matcher(s);
        // Loop removes matching nested parentheses from inside out.
        while (m.find()) {
            s = m.replaceFirst(m.group(1));
            m.reset(s);
        }
        regex = Pattern.compile("[()]");
        m = regex.matcher(s);
        // Check if there are any erroneous parentheses left over.
        if (m.find()) {
            return false;   // String has unbalanced parens.
        }
        return true;        // String has balanced parens.
    }
}

This does not validate the grammar, but may be useful as a first test to filter out obviously bad strings.

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各空 2024-11-10 08:25:14

可能有点过头了，但您可以考虑使用成熟的解析器生成器，例如 ANTLR (http://www.antlr.org/)。使用 ANTLR，您可以创建规则来自动生成 java 代码。假设您在输入中仅获得有效字符，这是一个语法分析问题，否则您需要首先使用词法分析来验证字符流。

对于语法分析，您可能有这样的规则：

PLUS : '+' ;
etc...

expression:
         term ( ( PLUS | MINUS | MULTIPLY | DIVIDE )^ term )*
      ;
term:
    constant
  | OPENPAREN! expression CLOSEPAREN!
  ;

常量是整数/实数。如果 ANTLR 生成的解析器代码无法将输入与解析器规则相匹配，它将抛出异常，以便您可以确定代码是否有效。

Possibly overkill, but you might consider using a fully fledged parser generator such as ANTLR (http://www.antlr.org/). With ANTLR you can create rules that will generate the java code for you automatically. Assuming you have only got valid characters in the input this is a syntax analysis problem, otherwise you would want to validate the character stream with lexical analysis first.

For syntax analysis you might have rules like:

PLUS : '+' ;
etc...

expression:
         term ( ( PLUS | MINUS | MULTIPLY | DIVIDE )^ term )*
      ;
term:
    constant
  | OPENPAREN! expression CLOSEPAREN!
  ;

With constant being integers/reals whatever. If the ANTLR generated parser code can't match the input with your parser rules it will throw an exception so you can determine whether code is valid.

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