Question

所以我已经编程了一段时间，而且我对程序的重要性感到困惑。格式良好＆＃34;我写了这个简单的公式程序。怎么可能＆＃34;更好＆＃34;写的？在程序中，您可以为爱因斯坦时间膨胀方程输入一些数字，并输出在一定时间后扩张了多少时间。

 import java.util.Scanner;
 import java.lang.Math;

public class Yass {
Scanner sc = new Scanner(System.in);
static double velocity;
static double time;
static double dilated_time;
static double dilated_time_Minutes;
static double dilated_time_Hours;
static double dilated_time_Days;
static double dilated_time_Years;
static boolean timeIsSeconds = false;
static boolean timeIsMinutes = false;
static boolean timeIsHours = false;
static boolean timeIsDays = false;
static boolean timeIsYears = false;
public static final double c = 299792.458; // km/s


//Time in seconds, velocity in kilometers/second

public Yass(){

    getInput();
    executeEquation(time, velocity, c);

}


public void executeEquation(double t, double v, double c) {

    dilated_time = t / (Math.sqrt(1 - (Math.pow(v, 2) / Math.pow(c, 2))));

    if(dilated_time <= 60){
        timeIsSeconds = true;
    }
    if(dilated_time > 60 && dilated_time < (60*60)){
        dilated_time_Minutes = dilated_time / 60;
        timeIsMinutes = true;
    }
    if(dilated_time > (60*60) && dilated_time < (60*60)*24){
        dilated_time_Hours = (dilated_time / 60) / 60;
        timeIsHours = true;
    }
    if(dilated_time > (60*60)*24 && dilated_time < ((60*60)*24)*365){
        dilated_time_Days = ((dilated_time / 60) / 60) / 24;
        timeIsDays = true;
    }
    if(dilated_time > ((60*60)*24)*365){
        dilated_time_Years = (((dilated_time / 60) / 60) / 24) / 365;
        timeIsYears = true;
    }
}


public void getInput(){
    System.out.println("Enter the time you want to measure(stationary) in seconds: ");
    time = sc.nextDouble();
    System.out.println("\n Enter velocity in km/s: ");
    velocity = sc.nextDouble();
}


public static void main(String[] args){

    new Yass();

    if(timeIsSeconds){
        System.out.println("So if you're in a spaceship and travel at " + velocity + 
                " km/s. And start a clock who'll measure " + time + " seconds.\nBut at the same time a stationary spaceship start an exact same clock"
                        + " at the same time and measure\nthe same amount of time. After the moving spaceship have measured " + time + " seconds. Then "
                                + "the stationary\nspaceship will have measured " + dilated_time + " seconds. And therefore time has been running slower for the moving spaceship.");
    }
    if(timeIsMinutes){
        System.out.println("So if you're in a spaceship and travel at " + velocity + 
                " km/s. And start a clock who'll measure " + time + " seconds.\nBut at the same time a stationary spaceship start an exact same clock"
                        + " at the same time and measure\nthe same amount of time. After the moving spaceship have measured " + time + " seconds. Then "
                                + "the stationary\nspaceship will have measured " + dilated_time_Minutes + " minutes. And therefore time has been running slower for the moving spaceship.");
    }
    if(timeIsHours){
        System.out.println("So if you're in a spaceship and travel at " + velocity + 
                " km/s. And start a clock who'll measure " + time + " seconds.\nBut at the same time a stationary spaceship start an exact same clock"
                        + " at the same time and measure\nthe same amount of time. After the moving spaceship have measured " + time + " seconds. Then "
                                + "the stationary\nspaceship will have measured " + dilated_time_Hours + " hours. And therefore time has been running slower for the moving spaceship.");
    }
    if(timeIsDays){
        System.out.println("So if you're in a spaceship and travel at " + velocity + 
                " km/s. And start a clock who'll measure " + time + " seconds.\nBut at the same time a stationary spaceship start an exact same clock"
                        + " at the same time and measure\nthe same amount of time. After the moving spaceship have measured " + time + " seconds. Then "
                                + "the stationary\nspaceship will have measured " + dilated_time_Days + " days. And therefore time has been running slower for the moving spaceship.");
    }
    if(timeIsYears){
        System.out.println("So if you're in a spaceship and travel at " + velocity + 
                " km/s. And start a clock who'll measure " + time + " seconds.\nBut at the same time a stationary spaceship start an exact same clock"
                        + " at the same time and measure\nthe same amount of time. After the moving spaceship have measured " + time + " seconds. Then "
                                + "the stationary\nspaceship will have measured " + dilated_time_Years + " years. And therefore time has been running slower for the moving spaceship.");
    }
}
}

示例：

以秒为单位输入您要测量的时间（静止）： 10

以km / s输入速度： 270000

所以如果你是一艘宇宙飞船并以270000.0公里/秒的速度旅行。并启动一个时间为10.0秒的时钟。但与此同时，固定的宇宙飞船同时启动一个完全相同的时钟并进行测量相同的时间。移动的宇宙飞船测得10.0秒后。然后静止宇宙飞船将测量23.009606245564093秒。因此，移动太空船的时间变得越来越慢。

Answer 1

突然出现一些事情。

使C平方不变（即静态最终），而不是每次计算它。
不接受“c”作为执行方程式方法的输入
制作一个单独的方法，该方法在几秒钟内获取一个长参数，并以最大的适用单位返回一个描述性字符串（例如，1000000 =＆gt;“11.57天”）。在此函数中，首先检查年份，因此在后续检查中，您只需要检查参数是否大于数字而不是两个数字之间。
摆脱main方法中的一系列“if”语句，只有一个语句在上面的＃3中调用新函数。

Answer 2

首先是使用javadoc提高可读性：

    /**
     * Calculates the result of Einsteins time dilation equation. 
     * @param t time in seconds
     * @param v speed in km/s
     * @param c light speed constant
     */
    public void executeEquation(double t, double v, double c) {

    }

在eclipse中使用格式化程序进行正确识别（Ctrl-Shift-F）。
变量v通常具有单位m / s。如果您使用kilomter / s，您可以重命名它：double vKmPerSec。应该对所有不是的单位进行此操作在SI（国际系统）单位（米，秒，公斤）。
＆＃34;亚斯＆＃34;对我来说不是一个明显的类名。将其重命名为更具描述性的内容。
类Yass没有评论，您可以将自己添加为作者
结果不应存储在静态变量中。＆＃34; dilated_time＆＃34;

最好使用结果类：

class EinsteinResult {
         // here you store dilated_time, etc.
}

并将结果返回

EinsteinResult executeEquation(double t, double v, double c) {
      //... 
      return EinsteinResult;
}

正如Peter Lawrey评论的那样，将Math.pow(c,2)替换为c*c

优化一个简单的程序

2 个答案: