java如何使用多个静态电源？

Question

尝试制作一个可以使用sumDigits、reverse和isPalindrome的程序。我需要打电话给他们，测试他们。我被告知使用公共静态int-sumdights（int-n）、公共静态int-reverse（int-number）和公共静态布尔值isPalindrome（int-number）。我只是很难让这三个程序一起运行。也不知道如何打印这三个问题的答案。任何和所有的帮助都将不胜感激。老实说，我现在能做的就是x.x.我很困惑

import java.util.Scanner;

public class Project3 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);

System.out.println("Enter an integer");
int n = input.nextInt();
int sum = sumDigits(n);
System.out.println("The sum is " + sum);
System.out.println("The reverse is " + n);
}

public static int sumDigits(long n) {
int num = (int)(n);
int sum = 0;

while (num > 0) {

sum += num % 10;
num = num / 10;

}
return sum;
}
public static int reverse(int number) {
int reverse = 0;
int rem = 0;
while (number != 0) {
rem = number % 10;
reverse = (reverse * 10) + rem;
number = number / 10;
}
return reverse;
}

public static boolean isPalindrome(int number) {
int reverse = reverse(number);
if(reverse == number) {
return true;
}
else 
return false;
}
}

Answer 1

要调用这三种方法，请使用main方法的名称，例如在main方法中：

public static void main(String arg[]){

   Scanner input = new Scanner(System.in);

   System.out.println("Enter an integer");
   int n = input.nextInt();
   int sum = sumDigits(n); //here is were the your first method is called
   System.out.println("The sum is " + sum);
   System.out.println("The reverse is " + n);

  int returnValue = reverse(5); //here is your second method call
  boolean rValue = isPalindrome(5); //here is your third method is called
}

我希望这有帮助

Answer 2

Java有静态和非静态的方法和变量

静电干扰

• 在运行前定义

• 如果在一个类中，它在所有实例之间共享

• 通常用于实用函数和常量

非静态的是

• 在运行时之前不可定义，需要实例化

• 我们称之为“普通”的方法，由单个类实例使用

现在在您的例子中，因为所有的都是静态的，所以可以从静态上下文（比如main）调用它们，而无需先实例化一个类

下面是一些使用静态和非静态方法的示例代码

public class Util{

    public static void main(String[] args){
        Util.method1("static"); // calls on the utils static method and prints "static"
        String s = Util.method2("static2"); // calls the method, and gets the string back
        System.out.println(s); // writes out the string it got back
        Util util = new Util(); // instantiates Util class to call non static methods
        util.method3(); // prints "not static", notice this is a call on the instance not the class

        System.out.println(isPalindrome(12321));
        System.out.println(Util.isPalindrome(123321));
        System.out.println(Util.isPalindrome(123456));
        System.out.println(util.isPalindrome(82218615));
        System.out.println(util.isPalindrome(0));
        System.out.println(util.isPalindrome(-12321));
        System.out.println(Util.isPalindrome(-123321));
        System.out.println(isPalindrome(-123456));
        System.out.println(util.isPalindrome(-82218615));
        // notice how a static can be acced in 3 ways
        // defined in this class, so no need to define where it's from.
        // call it on the class it's from, by calling on the class that contains it
        // call on an instance of the class that contains it
    }
    public static void method1(String input){
        System.out.println(input); // prints input
    }

    public static String method2(String input){
        return input; // just for the example, normally useless
    }

    public void method3(){ // notice not static
        System.out.println("not static");
    }

    public static boolean isPalindrome(int in){ // checks if palindrome
        String input = Integer.toString(in); // String of the input
        StringBuilder sr = new StringBuilder(); // StringBuilder to reverse one side
        if(input.charAt(0) == ('-')) // remove - from negative numbers assuming that is still supposed to pass the test
            input = input.substring(1, input.length());
        String left; // left side
        String right; // right side

        if(input.length() % 2 == 0){ // even lengths
            left = input.substring(0, input.length()/2); // adds in whole left side to left string
            right = sr.append(input.substring(input.length()/2, input.length())).reverse().toString();// adds in whole right side to right string
            // note that the function above used the reverse on the string side
        }
        else { // odd lengths
            left = input.substring(0, input.length()/2);// adds in whole left side to left string
            right = sr.append(input.substring(input.length()/2+1, input.length()).toString()).reverse().toString();// adds in whole right side to right string
            // note that the function above used the reverse on the string side
        }
        return right.equals(left);// returns if true or false (palindrome or not based on if right == left after reversed one side)
    }
}

这会产生这个输出（来自main方法）

static

static2

not static

true

true

false

false

true

true

true

false

false

Extra在上面添加了代码

如何查看isPalindrome 理论上，回文是像“12321”这样的东西，两种读法都是一样的，因此检查它们需要两件事

• 首先检查长度是否不均匀，如果不均匀，忽略中间部分，否则均匀分成两部分
• 第二，反转其中一边，然后比较两边是否相等

最后请注意，OOP、值数据封装和数据完整性，这就是为什么类通常处理自己的数据，并且只提供静态方法作为实用工具