下面是GitHub上另一个名为EvalEx的开源库。

与JavaScript引擎不同，这个库只专注于计算数学表达式。此外，该库是可扩展的，支持使用布尔运算符和圆括号。

我写了算术表达式的eval方法来回答这个问题。它可以做加法、减法、乘法、除法、求幂(使用^符号)，以及一些基本函数，如平方根。它支持使用(…)进行分组，并获得正确的操作符优先级和结合规则。

public static double eval(final String str) {
    return new Object() {
        int pos = -1, ch;
        
        void nextChar() {
            ch = (++pos < str.length()) ? str.charAt(pos) : -1;
        }
        
        boolean eat(int charToEat) {
            while (ch == ' ') nextChar();
            if (ch == charToEat) {
                nextChar();
                return true;
            }
            return false;
        }
        
        double parse() {
            nextChar();
            double x = parseExpression();
            if (pos < str.length()) throw new RuntimeException("Unexpected: " + (char)ch);
            return x;
        }
        
        // Grammar:
        // expression = term | expression `+` term | expression `-` term
        // term = factor | term `*` factor | term `/` factor
        // factor = `+` factor | `-` factor | `(` expression `)` | number
        //        | functionName `(` expression `)` | functionName factor
        //        | factor `^` factor
        
        double parseExpression() {
            double x = parseTerm();
            for (;;) {
                if      (eat('+')) x += parseTerm(); // addition
                else if (eat('-')) x -= parseTerm(); // subtraction
                else return x;
            }
        }
        
        double parseTerm() {
            double x = parseFactor();
            for (;;) {
                if      (eat('*')) x *= parseFactor(); // multiplication
                else if (eat('/')) x /= parseFactor(); // division
                else return x;
            }
        }
        
        double parseFactor() {
            if (eat('+')) return +parseFactor(); // unary plus
            if (eat('-')) return -parseFactor(); // unary minus
            
            double x;
            int startPos = this.pos;
            if (eat('(')) { // parentheses
                x = parseExpression();
                if (!eat(')')) throw new RuntimeException("Missing ')'");
            } else if ((ch >= '0' && ch <= '9') || ch == '.') { // numbers
                while ((ch >= '0' && ch <= '9') || ch == '.') nextChar();
                x = Double.parseDouble(str.substring(startPos, this.pos));
            } else if (ch >= 'a' && ch <= 'z') { // functions
                while (ch >= 'a' && ch <= 'z') nextChar();
                String func = str.substring(startPos, this.pos);
                if (eat('(')) {
                    x = parseExpression();
                    if (!eat(')')) throw new RuntimeException("Missing ')' after argument to " + func);
                } else {
                    x = parseFactor();
                }
                if (func.equals("sqrt")) x = Math.sqrt(x);
                else if (func.equals("sin")) x = Math.sin(Math.toRadians(x));
                else if (func.equals("cos")) x = Math.cos(Math.toRadians(x));
                else if (func.equals("tan")) x = Math.tan(Math.toRadians(x));
                else throw new RuntimeException("Unknown function: " + func);
            } else {
                throw new RuntimeException("Unexpected: " + (char)ch);
            }
            
            if (eat('^')) x = Math.pow(x, parseFactor()); // exponentiation
            
            return x;
        }
    }.parse();
}

例子:

System.out.println(eval("((4 - 2^3 + 1) * -sqrt(3*3+4*4)) / 2"));

输出:7.5(正确)

该解析器是递归下降解析器，因此在内部为其语法中的每个操作符优先级使用单独的解析方法。我故意保持简短，但以下是一些你可能想要扩展的想法:

Variables: The bit of the parser that reads the names for functions can easily be changed to handle custom variables too, by looking up names in a variable table passed to the eval method, such as a Map<String,Double> variables. Separate compilation and evaluation: What if, having added support for variables, you wanted to evaluate the same expression millions of times with changed variables, without parsing it every time? It's possible. First define an interface to use to evaluate the precompiled expression: @FunctionalInterface interface Expression { double eval(); } Now to rework the original "eval" function into a "parse" function, change all the methods that return doubles, so instead they return an instance of that interface. Java 8's lambda syntax works well for this. Example of one of the changed methods: Expression parseExpression() { Expression x = parseTerm(); for (;;) { if (eat('+')) { // addition Expression a = x, b = parseTerm(); x = (() -> a.eval() + b.eval()); } else if (eat('-')) { // subtraction Expression a = x, b = parseTerm(); x = (() -> a.eval() - b.eval()); } else { return x; } } } That builds a recursive tree of Expression objects representing the compiled expression (an abstract syntax tree). Then you can compile it once and evaluate it repeatedly with different values: public static void main(String[] args) { Map<String,Double> variables = new HashMap<>(); Expression exp = parse("x^2 - x + 2", variables); for (double x = -20; x <= +20; x++) { variables.put("x", x); System.out.println(x + " => " + exp.eval()); } } Different datatypes: Instead of double, you could change the evaluator to use something more powerful like BigDecimal, or a class that implements complex numbers, or rational numbers (fractions). You could even use Object, allowing some mix of datatypes in expressions, just like a real programming language. :)

这个答案中的所有代码都已发布到公共领域。玩得开心!

一个可以计算数学表达式的Java类:

package test;

public class Calculator {

    public static Double calculate(String expression){
        if (expression == null || expression.length() == 0) {
            return null;
        }
        return calc(expression.replace(" ", ""));
    }
    public static Double calc(String expression) {
        String[] containerArr = new String[]{expression};
        double leftVal = getNextOperand(containerArr);
        expression = containerArr[0];
        if (expression.length() == 0) {
            return leftVal;
        }
        char operator = expression.charAt(0);
        expression = expression.substring(1);

        while (operator == '*' || operator == '/') {
            containerArr[0] = expression;
            double rightVal = getNextOperand(containerArr);
            expression = containerArr[0];
            if (operator == '*') {
                leftVal = leftVal * rightVal;
            } else {
                leftVal = leftVal / rightVal;
            }
            if (expression.length() > 0) {
                operator = expression.charAt(0);
                expression = expression.substring(1);
            } else {
                return leftVal;
            }
        }
        if (operator == '+') {
            return leftVal + calc(expression);
        } else {
            return leftVal - calc(expression);
        }

    }
    
    private static double getNextOperand(String[] exp){
        double res;
        if (exp[0].startsWith("(")) {
            int open = 1;
            int i = 1;
            while (open != 0) {
                if (exp[0].charAt(i) == '(') {
                    open++;
                } else if (exp[0].charAt(i) == ')') {
                    open--;
                }
                i++;
            }
            res = calc(exp[0].substring(1, i - 1));
            exp[0] = exp[0].substring(i);
        } else {
            int i = 1;
            if (exp[0].charAt(0) == '-') {
                i++;
            }
            while (exp[0].length() > i && isNumber((int) exp[0].charAt(i))) {
                i++;
            }
            res = Double.parseDouble(exp[0].substring(0, i));
            exp[0] = exp[0].substring(i);
        }
        return res;
    }


    private static boolean isNumber(int c) {
        int zero = (int) '0';
        int nine = (int) '9';
        return (c >= zero && c <= nine) || c =='.';
    }

    public static void main(String[] args) {
        System.out.println(calculate("(((( -6 )))) * 9 * -1"));
        System.out.println(calc("(-5.2+-5*-5*((5/4+2)))"));

    }

}

这是另一个有趣的选择 https://github.com/Shy-Ta/expression-evaluator-demo

它的用法非常简单，可以完成工作，例如:

  ExpressionsEvaluator evalExpr = ExpressionsFactory.create("2+3*4-6/2");  
  assertEquals(BigDecimal.valueOf(11), evalExpr.eval()); 

你可以看看Symja框架:

ExprEvaluator util = new ExprEvaluator(); 
IExpr result = util.evaluate("10-40");
System.out.println(result.toString()); // -> "-30" 

请注意，可以计算更复杂的表达式:

// D(...) gives the derivative of the function Sin(x)*Cos(x)
IAST function = D(Times(Sin(x), Cos(x)), x);
IExpr result = util.evaluate(function);
// print: Cos(x)^2-Sin(x)^2

解决这个问题的正确方法是使用词法分析器和解析器。您可以自己编写这些页面的简单版本，或者这些页面还包含指向Java词法分析器和解析器的链接。

创建递归下降解析器是非常好的学习练习。