如何在Java中计算结果？

Question

我创建这段代码是为了创建随机数和运算符，但是它如何计算和显示结果呢？

例如，打印输出: 4+1-3或9*2-8等，我不知道如何计算4+1-3或9*2-8的结果并打印出来。

public static void main(String[] args) {
    int t = 0;
    String[] operators = {"-", "+", "*"};
    String operator;
    Random r = new Random();

    for (int i = 0; i < 3; i++) {
        int randNum = randNums(9, 1);
        operator = operators[r.nextInt(operators.length)];
        System.out.print(randNum);
        if (t < 2) {
            System.out.print(operator);
            t++;
        }
    }

}

Answer 1

这是一些(相对)简单的代码，它从左到右对表达式进行will calculate (它没有考虑操作的顺序，所以3+4*5被计算为(3+4)*5，而不是正确的3+(4*5))：

public static void main(String[] args) {
    String[] operators = {"-", "+", "*"};
    String operator;
    Random r = new Random();
    int result = -1;

    for (int i = 0; i < 3; i++) {
        int randNum = r.nextInt(9) + 1; // 1 - 9 inclusive
        if (i != 0) {
            operator = operators[r.nextInt(operators.length)];
            System.out.print(operator);
            result = calculate(result, randNum, operator);
        }
        else {
            result = randNum;
        }
        System.out.print(randNum);
    }
    System.out.println("=" + result);
}
public static int calculate(int operand1, int operand2, String operator) {
    switch (operator) {
        case "+":
            return operand1 + operand2;
        case "-":
            return operand1 - operand2;
        case "*":
            return operand1 * operand2;
        default:
            throw new RuntimeException();
    }
}

Answer 2

这不是一个简单的问题，因为你必须牢记算术运算符的优先级，所以我猜你必须使用一个对你有帮助的数学库。例如，Formula4j：http://www.formula4j.com/index.html

Answer 3

您是否坚持将操作应用于参数？

最简单的方法：

if ("+".equals(operator)) {
  result = arg1 + arg2;
} else if ...
System.out.println("See, " + arg1 + operator + arg2 " = " + result);

一种更具可扩展性的方式，它使用哈希表而不是无限的if

import java.util.*;

// Poor man's first-class function
interface BinaryOp {
  int apply(int arg1, int arg2);
}

final static BinaryOp ADD = new BinaryOp() {
  public int apply(int arg1, int arg2) { return arg1 + arg2; }
}

final static BinaryOp SUBTRACT = new BinaryOp() {
  public int apply(int arg1, int arg2) { return arg1 - arg2; }
}

final static BinaryOp MULTIPLY = new BinaryOp() {
  public int apply(int arg1, int arg2) { return arg1 * arg2; }
}

static final Map<String, BinaryOp> OPERATIONS = new HashMap<String, BinaryOp>();

// This replaces the 'if', easier to extend.
static {
  OPERATIONS.put("+", ADD);
  OPERATIONS.put("-", SUBTRACT);
  OPERATIONS.put("*", MULTIPLY);
}

public static void main(String[] args) {
  ...
  BinaryOp operation = OPERATIONS.get(operation_name);
  int result = operation.apply(arg1, arg2);
  ...
}

如果你认为这是不必要的冗长，它是。类似这样的东西仍然是Java-land中的典型模式。(这就是Scala存在的原因，但这是另一回事。)

这甚至没有触及操作优先级或括号。