Question

我正在尝试在分数之间创建一个计算器，如果用户输入+， - ，*或/它将在该情况下对应。这是我到目前为止的代码：

import java.util.Scanner;
public class calculator 
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        String x,y;


        System.out.println("Enter first fraction in a / b form: ");
        x = input.nextLine();

        System.out.println("Enter operation: ");
        char z = input.next().charAt(0);

        System.out.println("Enter second fraction in c / d form: ");
        y = input.nextLine();

        String aString = x.substring(0,1);
        String bString = x.substring(4,5);
        String cString = x.substring(0,1);
        String dString = x.substring(4,5);

        int a = Integer.parseInt(aString);
        int b = Integer.parseInt(bString);
        int c = Integer.parseInt(cString);
        int d = Integer.parseInt(dString);


        int answer = 0;
        switch (z)
        {
            case '+':
                answer = (a/b) + (c/d);
                break; 
            case '-':
                answer = (a/b) - (c/d);
                break;
            case '*':
                answer = (a/b) * (c/d); 
                break;
            case '/':
                answer = (a/b) /(c/d);
                break;
            default:
                System.out.println("ERROR");
                break;              
        }
        System.out.println("Answer = " + answer);       
    }
}

输出应该

Enter first fraction in a / b form: 
1 / 2
Enter operation: 
+
Enter second fraction in c / d form: 
2 / 5
answer = 9/10

Answer 1

问题出在这里

 String aString = x.substring(0,1);
 String bString = x.substring(4,5);
 String cString = x.substring(0,1);
 String dString = x.substring(4,5);

a=c和b=d

这两行

 String cString = x.substring(0,1);
 String dString = x.substring(4,5);

应该是

String cString = y.substring(0,1);
 String dString = y.substring(4,5);

在switch打印a b c d z的值之前，您将了解自己的正确性。

还有一件事是，您正在对int个变量进行划分，因此结果只会在int中。

我建议您将a b c d和answer的类型更改为double并使用Double.parseDouble()将string转换为double。

Answer 2

这个问题非常适合OOP，一个额外的类：

public class Q {

    public final int numerator;
    public final int denominator;

    public static Q valueOf(String representation) {
        Pattern pattern = Pattern.compile("([-]?\\d+) *[/:] *([-]?\\d+)");
        Matcher matcher = pattern.matcher(representation.trim());
        if (!matcher.matches()) {
            throw new IllegalArgumentException(
                "Not a quotient (like '3/4'): " + representation);
        }
        int num = Integer.parseInt(matcher.group(1));
        int den = Integer.parseInt(matcher.group(2));
        return new Q(num, den);
    }

    public Q(int num, int den) {
        if (den < 0) {
            den = -den;
            num = -num;
        }
        int c = gcd(Math.abs(den), Math.abs(num));
        denominator = den / c;
        numerator = num / c;
    }

    @Override
    public String toString() {
        return numerator + "/" + denominator;
    }

    public Q add(Q rhs) {
        int c = gcd(denominator, rhs.denominator);
        int den = (denominator / c) * rhs.denominator;
        int num = numerator * (rhs.denominator / c)
                + rhs.numerator * (denominator / c);
        return new Q(num, den);
    }

    public Q mult(Q rhs) {
        int den = denominator * rhs.denominator;
        int num = numerator * rhs.numerator;
        return new Q(num, den);
    }

    public static int gcd(int x, int y) {
        assert x >= 0 && y >= 0;
        while (x != y) {
            if (x > y) {
                x -= y;
            } else {
                y -= x;
            }
        }
        return x;
    }
}

使用这个类，几个有问题的小代码部分就会消失。

输入一个特殊字符，然后在开关盒中使用

2 个答案: