Question

我这里有一个简单的计算器，带有一个switch语句，允许用户输入一个字符串，如下所示： * 5

运营商可以是 - + /或* 代码在添加期间正确执行，但是，当我减去一个值时，它会添加它，并且除以或乘以将导致下面列出的例外。

Enter an operator and a number:
+5
Enter an operator and a number:
*2
Exception in thread "main" java.lang.NumberFormatException: For input string: "*2"
    at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:2043)
    at sun.misc.FloatingDecimal.parseDouble(FloatingDecimal.java:110)
    at java.lang.Double.parseDouble(Double.java:538)
    at Calculator.whatOperator(Calculator.java:38)
    at Calculator.aResult(Calculator.java:29)
    at Main.main(Main.java:30)

我相信我没有正确解析双重因为某些原因但我不确定如何这样做...我在想stringtokenizer但是如何在没有分隔符的情况下使用tokenizer？这是我的计算器课程，下面是我的主要课程

import java.util.Scanner;
public class Calculator {
    private final int RESET = 0;
    private double number = 0;
    private double result = 0; 
    private char operator;
    private Scanner keyboard = new Scanner(System.in);
    public Calculator(double number)
    {
        this.number = number;

    }
    public void reset()
    {
        this.number = RESET;
    }
    public double aResult(Calculator other)
    {

        other.whatOperator();
        this.result = other.result;
        return result;

    } 

    public void whatOperator()
    {
        String operatorString = enterNumber();
        // the error occurs here....is there a better way to do this?
        double theNumber = Double.parseDouble(operatorString);
        char theOperator = operatorString.charAt(0);
        this.operator = theOperator;
        operatorString ="";
        operatorString += theOperator;

        // the switch should perform the operation 
        switch(operatorString){
        case "*":
        result = getNumber() * theNumber;
        break;
        case "/":
        result = getNumber() / theNumber;
        break;
        case "+":
        result = getNumber() + theNumber;
        break;
        case "-":
        result = getNumber() - theNumber;
        break;
        case "R":
        result = RESET;
        break;
        case "P":
        System.out.println("Goodbye");
        System.exit(0);

    }


}
public double add(double secondNumber)
{
    result = number + secondNumber;
    return result;

}
public double divide(double secondNumber)
{
    result = number / secondNumber;
    return result;
}
public double multiply(double secondNumber)
{
    result = number * secondNumber;
    return result;
}
public void subtract(double secondNumber)
{
    result = number - secondNumber;
}


public double getNumber()
{
    return number;
}
public void setNumber(double number)
{
    this.number = number;
}
    public  String enterNumber()
    {

        System.out.println("Enter an operator and a number:");
        String toString = keyboard.nextLine();
        return toString;
    }

}
public class Main {

    public static void main (String[] args) {
        Calculator a = new Calculator(0);
        a.setNumber(a.aResult(a));
        a.setNumber(a.aResult(a));
        String theString = String.valueOf(a.getNumber());
        System.out.println(theString);
    }
}

Answer 1

这一行错了：

double theNumber = Double.parseDouble(operatorString);

因为它将操作符解析为数字的一部分。相反，使用这个：

double theNumber = Double.parseDouble(operatorString.substring(1));

为什么不解析Double表现正确？

1 个答案: