缩短超大数字的可能方法？

Question

这是我程序中双“m”的值（经过计算后的行星质量，这个值是专门针对地球质量的）

• 5.9734054370474745E24

使用System.out.println（m）打印时；输出为

• 5.9734054370474745E24（正确输出）

使用系统打印时。出来println（数学圆（m））；输出是

• 9223372036854775807（输出错误）

我怎样才能缩短m的值，使其在%6s之内？比如说

• 5.97E24

下面是我的代码。作业要求我们在格式化的图表中输出最终值（正如我在下面尝试的那样）

//Java imports
import java.util.*;
import java.lang.Math;

//Main class
class Main {
  public static void main(String[] args) {
    //Initializing scanner name userInput
    Scanner userInput = new Scanner (System.in);

    //Greeting statement, prompts user to input the circumference in km
    System.out.println("\nWelcome to the Escape Velocity Application. To begin, please enter the following information below. \nEnter the circumference (km):");

    //Stores circumference in double named "circum" and converts the value to its meter equivalent (unit conversion required)
    double circum = userInput.nextDouble() * Math.pow(10, 3); 

    //Prompts user to input the acceleration in m/s^2
    System.out.println("Enter the acceleration due to gravity (m/s^2):");

    //Stored value in double named "f"
    double f = userInput.nextDouble(); 

    //Gravitational Constant
    double G = 6.67408e-11;

    //1 - Radius calculation using the circumference of a circle formula
    double r = circum/(2*Math.PI);

    //2 - Mass calculation using the gravity formula
    double m = f*Math.pow(r, 2)/G;

    //3 - Calculation escape velocity using the escape velocity formula
    double e = (Math.sqrt((2.0*G*(m))/r));

    //Final output statements
    System.out.println("\nThe radius is: " + Math.round(r * Math.pow(10, -3)) + " kilometers.");
    System.out.println("The mass is: " + m + " kg.");
    System.out.println("The escape velocity is: " + Math.round(e) + " m/s.");

    //Formatted output statements
    System.out.format("\n%20s %6s %10s", "Radius:", Math.round(r * Math.pow(10, -3)), "km.");
    System.out.format("\n%20s %6s %10s", "Mass:", Math.round(m), "kg.");
    System.out.format("\n%20s %6s %10s", "Escape Velocity:", Math.round(e), "m/s.");
  }
}

这就是输出的样子。由于m的长值，第二条直线的中心偏移。

The radius is: 6378 kilometers.
The mass is: 5.973405437304745E24 kg.
The escape velocity is: 11181 m/s.

         Radius:   6378        km.
           Mass: 9223372036854775807        kg.
Escape Velocity:  11181       m/s.

Answer 1

您可以使用以下代码：

double x = 5.973405437304745e24;
System.out.printf("Mass: %.2e kg.%n", x);

输出<代码>质量：5.97e 24 kg 。

%.2e格式化数字，而%n只是添加了一个换行符。.2指定点后需要两位小数。e要求格式化程序使用科学符号。

数学的问题。round（）的意思是，结果存储在一个不能代表如此大的数字的long中。