The question of passing Java values is illustrated in an intuitive way. If you have a high opinion, please make an axe!

For the differences between the two code segments and the results, use the code to illustrate the memory and execution process.
Thank you!

    public class PassAddr {          public static void main(String[] args) {              String s=new String("old");  //1            method(s);  //3            System.out.println(s);          }          static void method(String str){              str=new String("new");  //2        }      }  

The output result is: Old.

Public class t {public static void main (string [] ARGs) {string [] arr = new string [2]; arr [0] = "old_0 "; arr [1] = "old_1"; // 11/* arr [0] = new string ("old_0"); arr [1] = new string ("old_1 "); // 11 similarly */method (ARR); // 14 system. out. println (ARR [0] + ";" + arr [1]);} static void method (string [] A) {// 12 A [0] = "new_0 "; A [1] = "new_1"; // 13 }}

 

This output is: new_0; new_1
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I thought about the problem for a long time. Next, I hope you will be right!
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String: // at 1:

// 2:

// 3:

Because the method call has been completed, the STR Temporary Variable disappears in the stack, S is still 0x001, and the object new string ("old") at the address is not passive; the output result is old;

--------------------------------------------------- Stubborn split line ------------------------------------------------------

// Memory status at 11:

// 12 points:

// Memory status at 13:

At this time, the values of a [0] And a [1] are changed, which is actually the corresponding value of arr. Finally, the result is displayed. // 14

So the result changes.

This plot is intuitive.