Question

这是我用Java编写的基本计算器。显示输出后，我想问用户是否要继续，如果他回答“是”，则重复该过程。我已经检查了许多有关堆栈溢出的主题，但我仍然不知道如何通过 do while循环实现我的目标，任何人都可以帮忙吗？

import java.util.Scanner;

public class SecondQuestion {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        double firstN; 
        double secondN; 
        char operator;

        Scanner readInput = new Scanner(System.in);

        System.out.printf("Type a number, operator, number --" + "separated by a space: "); 

        firstN = readInput.nextDouble();
        operator = readInput.next().charAt(0); 
        secondN = readInput.nextDouble();

        if (operator == '+')
            System.out.printf("%f + %f = %f", firstN, secondN, firstN + secondN); 

        else if (operator == '-') 
            System.out.printf("%f - %f = %f", firstN, secondN, firstN - secondN); 

        else if (operator == '*')
            System.out.printf("%f * %f = %f", firstN, secondN, firstN * secondN); 

        else if (operator == '/')
            System.out.printf("%f / %f = %f", firstN, secondN, firstN / secondN); 

        else if (operator == '%')
            System.out.printf("%f %% %f = %f", firstN, secondN,firstN % secondN); 

        else
            System.out.printf("Unknown operator"); 
        System.out.printf("\n\n");

        int loopCount = 0;
        char charResponse='y';
        Scanner readInput1 = new Scanner (System.in);
        while (charResponse !='y')
        {
            System.out.println("press y");
            charResponse = readInput1.next().charAt(0);
        }
    }

}

Answer 1

def path_mas_largo (labyrinth, idlaberinto):
    for i in range (len (labyrinth [idlaberinto]) - 1):
        if len (labyrinth [idlaberinto] [i] ['ensenanza' + str (i)]) > len (labyrinth [idlaberinto] [i + 1] ['ensenanza' + str (i + 1)]):
            j = len (labyrinth [idlaberinto] [i] ['ensenanza' + str (i)])
            if j > len (labyrinth [idlaberinto] [i + 1] ['ensenanza' + str (i + 1)]):
               camino_largo = laberinto [idlaberinto] [i] ['ensenanza' + str (i)]
               print (long_path)
               emular_recorrido (camino_largo, ventana, blanco, negro, celeste, aumento)

Answer 2

使用do-while：

public static void main(String[] args) {
    // TODO Auto-generated method stub

    double firstN;
    double secondN;
    char operator;

    Scanner readInput = new Scanner(System.in);
    char charResponse = 'y';
    do {
        System.out.printf("Type a number, operator, number --" + "separated by a space: ");

        firstN = readInput.nextDouble();
        operator = readInput.next().charAt(0);
        secondN = readInput.nextDouble();

        if (operator == '+')
            System.out.printf("%f + %f = %f", firstN, secondN, firstN + secondN);

        else if (operator == '-')
            System.out.printf("%f - %f = %f", firstN, secondN, firstN - secondN);

        else if (operator == '*')
            System.out.printf("%f * %f = %f", firstN, secondN, firstN * secondN);

        else if (operator == '/')
            System.out.printf("%f / %f = %f", firstN, secondN, firstN / secondN);

        else if (operator == '%')
            System.out.printf("%f %% %f = %f", firstN, secondN, firstN % secondN);

        else
            System.out.printf("Unknown operator");
        System.out.printf("\n\n");

        int loopCount = 0;
        Scanner readInput1 = new Scanner(System.in);
        System.out.println("press y to continue");
        charResponse = readInput1.next().charAt(0);
    } while (charResponse == 'y');

}

do-while循环是一个后递增循环，因此它将首先执行，然后在一段时间内检查条件。

如何使用do while询问用户是否要重复计算

2 个答案: