Question

我正在完成一项任务，虽然我已经完成了它的开始部分，但我仍然处于最后一步。这是我必须要做的最后一部分。

修改main方法，以便在调用swap方法之后， num1和num2的实际值在main中交换，并显示在此处输出。输出将有一个额外的行，显示交换后main中的两个数字的值。新线在输出中需要“在主方法中交换数字后， num1是2，num2是1“。

代码：

     public class PassByValue {
  /** Main method */
    public static void main(String[] args) {
        // Declare and initialize variables
        int num1 = 1;
        int num2 = 2;

        System.out.println("Before invoking the swap method, num1 is " +
        num1 + " and num2 is " + num2);

        //invoke the swap method to attempt to swap two variables
        swap(num1, num2);

        System.out.println("After invoking the swap method, num1 is " +
         num1 + " and num2 is " + num2);
    }

    /** Swap two variables */
    public static void swap(int n1, int n2) {
        System.out.println("\tInside the swap method");
        System.out.println("\t\tBefore swapping, n1 is " + n1
                + " and n2 is " +n2);

        // Swap n1 with n2
        int temp = n1;
        n1 = n2;
        n2 = temp;

        System.out.println("\t\tAfter swapping, n1 is " + n1 
                + " and n2 is " + n2);
    }
}

Answer 1

我将在这里向您展示两种方法，选择哪种更好。第一：

public class Swap{

      static int num1 = 1;//declare num1 and num2 as static
      static int num2 = 2;
    /** Swap two variables */
    public static void swap(int n1, int n2) {
        System.out.println("\tInside the swap method");
        System.out.println("\t\tBefore swapping, n1 is " + n1
                + " and n2 is " +n2);

        // Swap n1 with n2
        int temp = n1;
        n1 = n2;
        n2 = temp;

        System.out.println("\t\tAfter swapping, n1 is " + n1 
                + " and n2 is " + n2);
        num1=n1;//after swapping assign the new swapped values to num1 and num2. It will reflect in main as it is declared as static
        num2=n2;
    }

    public static void main(String[] args) throws UnknownHostException {
         // Declare and initialize variables


        System.out.println("Before invoking the swap method, num1 is " +
        num1 + " and num2 is " + num2);

        //invoke the swap method to attempt to swap two variables
        swap(num1, num2);

        System.out.println("After invoking the swap method, num1 is " +
         num1 + " and num2 is " + num2);
     }


}

方法2：

public class Swap{

      static int num1 = 1;
      static int num2 = 2;
    /** Swap two variables */
    public static void swap() {//without functional arguments
        System.out.println("\tInside the swap method");
        System.out.println("\t\tBefore swapping, n1 is " + num1
                + " and n2 is " +num2);

        // Swap n1 with n2
        int temp = num1;//<-- swap values here and it will reflect
        num1 = num2;
        num2 = temp;

        System.out.println("\t\tAfter swapping, n1 is " + num1 
                + " and n2 is " + num2);     
    }

    public static void main(String[] args) throws UnknownHostException {
         // Declare and initialize variables


        System.out.println("Before invoking the swap method, num1 is " +
        num1 + " and num2 is " + num2);

        //invoke the swap method to attempt to swap two variables
        swap();//dont pass arguments as num1 and num2 are declared static, swap the values in the method itself

        System.out.println("After invoking the swap method, num1 is " +
         num1 + " and num2 is " + num2);
     }


}

Answer 2

我认为练习的目的是向你展示这是不可能的，因为Java只有“按值传递”。要做这样的事情，你需要传递一个包含值的对象的swap方法。例如，传递一个数组，然后swap可以交换数组中的前两个值。或者定义一些其他对象。

Answer 3

你的教授可能已经完成了这项任务，你可以理解Pass by value并通过java中的参考概念和OOPS传递。

首先在java中，method参数将作为pass by value发送。如果参数是原始类型，则将发送它们的值。如果参数是“对象”，则对象的引用仍将作为值发送。

所以，在这里解决你的问题，你可以通过创建一个包含你的值的对象来解决它。

以下代码可以解决您的问题：

 public class test {
 /** Main method */
public static void main(String[] args) {
    // Declare and initialize variables

    Numberr num1 = new Numberr(1);
    Numberr num2 = new Numberr(2);


    System.out.println("Before invoking the swap method, num1 is " +
    num1.n + " and num2 is " + num2.n);

    //invoke the swap method to attempt to swap two variables
    swap(num1, num2);

    System.out.println("After invoking the swap method, num1 is " +
     num1.n + " and num2 is " + num2.n);
}

/** Swap two variables */
public static void swap(Numberr n1, Numberr n2) {
    System.out.println("\tInside the swap method");
    System.out.println("\t\tBefore swapping, n1 is " + n1.n
            + " and n2 is " +n2.n);

    // Swap n1 with n2
    int temp = n1.n;
    n1.n = n2.n;
    n2.n = temp;

    System.out.println("\t\tAfter swapping, n1 is " + n1 .n
            + " and n2 is " + n2.n);
}


}    
class Numberr {
public Numberr(int i) {
        // TODO Auto-generated constructor stub
    this.n = i;
}

int n;

  }

Java：通过Value，swap方法传递

3 个答案: