如何检查字符串是否仅含运算符？正则表达式非法字符范围报错排查

Hey there! Let's break down what's going wrong with your regex and get your operator validation logic sorted out.

1. Why the "Illegal character range" Error Happens #

That error pops up because in your regex [(+=-*/^)]+, the sequence =-* is being interpreted as a character range. The ASCII value of = is 61, and * is 42—so you're trying to define a range from a higher value to a lower one, which regex engines don't allow.

2. Fixing the Regex #

There are two simple ways to fix this:

• Move the - to the start or end of your character set, so it's not treated as a range operator: [(+*/^=-)]
• Escape the - with a backslash (remember, in Java strings you need double backslashes): "[(+\\-*/^=)]+"

3. Implementing the Full Operator-Only Check #

Your current code only checks if the stack top matches the regex, but let's make sure we properly validate that a string contains only your target operators (no extra characters). Here's a revised version of your code with the fix and improved logic:

import java.util.Stack;

public class StringConverter {
    private Stack<String> stringStack = new Stack<>();
    private Stack<Double> digitStack = new Stack<>();

    public void convertString(String s) {
        String[] arr = s.split("\"");
        for (String segment : arr) {
            stringStack.push(segment);
        }

        // Fixed regex: placed '-' at the end to avoid range issues
        String operatorRegex = "[(+*/^=-)]+";
        
        if (!stringStack.isEmpty()) {
            String topElement = stringStack.peek();
            if (topElement.matches("[0-9]+")) {
                digitStack.push(Double.valueOf(stringStack.pop()));
            } else if (topElement.matches(operatorRegex)) {
                // Add your operator handling logic here (e.g., push to operator stack)
                System.out.println("Top element is a valid operator string: " + topElement);
            } else {
                System.out.println("Top element is neither a digit nor valid operators");
            }
        }

        System.out.println("Digit stack contents: " + digitStack);
    }

    // Helper method to check if any string is made up of only your operators
    public boolean isOnlyOperators(String input) {
        if (input == null || input.isBlank()) {
            return false; // Adjust this if empty string should be considered valid
        }
        String operatorRegex = "[(+*/^=-)]+";
        return input.matches(operatorRegex);
    }
}

Quick Notes #

• The matches() method in Java automatically checks the entire string (it's equivalent to wrapping your regex in ^ and $), so you don't need to add those anchors manually.
• If you need to add/remove operators later, just update the characters inside the [] in the regex.
• Characters like +, *, / don't need escaping inside a regex character set—only - (when not at the edges), ^ (when at the start), [, and ] require special handling.

内容的提问来源于stack exchange，提问作者Dartweiler