Java中温度转换器的优化答案

【问题标题】：Optimization of Temperature converter in JavaJava中温度转换器的优化
【发布时间】：2020-09-22 17:46:35
【问题描述】：

所以我正在努力重新开始编程，我编写了这段代码，将摄氏度转换为华氏度，反之亦然。

它工作正常，但我必须让它更短。你能用更少的线让它工作吗？

public static void main(String[] args) {
    Double temperature = 0.0;
    Double result = 0.0;
    
    String menuChoise = "1. Celsius to Fahrenheit \n2. Fahrenheit to Celsius";
    int anyChoiseMenu = 3;

    System.out.println("Chose a conversion");
    System.out.println(menuChoise);

    Scanner sc = new Scanner(System.in);

    while (true) {
        try {
            anyChoiseMenu = sc.nextInt();
            if (anyChoiseMenu == 1) {
                System.out.println("What temperature you wanna convert: ");
                temperature = sc.nextDouble();
                result = (1.8) * temperature + 32;
                System.out.println("Result: " + result + " Fahrenheit");
                break;
            } else if (anyChoiseMenu == 2) {
                System.out.println("What temperature you wanna convert: ");
                temperature = sc.nextDouble();
                result = ((5 * (temperature - 32.0)) / 9.0);
                System.out.println("Result: " + result + " Celsius");
                break;
            } else {
                System.out.println("Wrong number. Try again!");
                sc.nextLine();
            }
        } catch (Exception e) {
            System.out.println("Only numbers bro.");
            System.out.println(menuChoise);
            sc.nextLine();
            continue;
        }
    }
}

【问题讨论】：

您好，欢迎您。是的，你可以用更少的行来写它，但为什么要这样做呢？这很好，在逻辑上建立起来，并且可读性很好。它也可以完成这项工作。优化您的开发人员同事的时间投资，这可能包括您未来的自己，而不是代码行数。如果当前性能可证明不令人满意，则仅针对计算机进行优化。
如果它有效，那么它应该在Code Review

标签： java optimization converters

【解决方案1】：

您的代码似乎运行良好。通过一些技巧，如下所示，您可以剪切几行代码。

import java.util.InputMismatchException;
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String menuChoice = "\nChoose a conversion: \n1. Celsius to Fahrenheit \n2. Fahrenheit to Celsius";
        int menu;

        while (true) {
            System.out.println(menuChoice);
            try {
                menu = sc.nextInt();
            } catch (InputMismatchException e){
                System.out.println("Invalid input, only numbers are allowed!");
                sc.nextLine();
                continue;
            }

            if (menu >= 3) {
                System.out.println("Wrong number. Try again!");
            } else {
                System.out.print("What temperature you wanna convert: ");
                double temperature = sc.nextDouble();
                if (menu == 1) {
                    temperature = 1.8 * temperature + 32;
                } else if (menu == 2) {
                    temperature = (temperature - 32) / 1.8;
                }
                String grade = (menu == 1)? "Celsius": "Fahrenheit";
                System.out.format("Result: %.2f " + grade + '\n', temperature);
                break;
            }
        }
    }
}

【讨论】：

【解决方案2】：

public static void main(String... args) {
    Scanner scan = new Scanner(System.in);
    scan.useLocale(Locale.ENGLISH);

    while (true) {
        System.out.println("Chose a conversion:");
        System.out.println("1. Celsius to Fahrenheit");
        System.out.println("2. Fahrenheit to Celsius");
        System.out.println("3. Exit");

        System.out.print("> ");
        int menu = scan.nextInt();

        if (menu == 1) {
            System.out.print("What temperature you wanna convert: ");
            double celsius = scan.nextDouble();
            double fahrenheit = 1.8 * celsius + 32;
            System.out.format(Locale.ENGLISH, "Result: %.2f Fahrenheit\n", fahrenheit);
        } else if (menu == 2) {
            System.out.println("What temperature you wanna convert: ");
            double fahrenheit = scan.nextDouble();
            double celsius = (fahrenheit - 32) / 1.8;
            System.out.format(Locale.ENGLISH, "Result: %.2f Celsius\n", celsius);
        } else if (menu == 3)
            break;
        else
            System.out.println("Wrong number. Try again!");

        System.out.println();
    }
}

【讨论】：