When Java passes a parameter, the difference between the value of the pass and the reference

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Learning Zhenghaipeng in PKU

When I recently reviewed the nine sorting algorithms, unexpected discoveries were made by bubbling sort, inserting sort, selecting sort, etc. in-place sort,

Sort (a);

The output of A is a new array already lined up.

However, sort (a) is used when sorting by Out-place and so on. After that, the output is still the original array!

Why is it? Please look at the code and the comments inside:

Package Zhp.outplace_sort;import ZHP. Tools;public class Test {public static void main (string[] args) {int[] A1 = {1, 2, 3, 4, 5};int[] A2 = {1, 2, 3, 4, 5};F1 ( A1); F2 (A2); Tools.print (A1); Output array tools.print (A2);} /** * Method Changes the value of the parameter (in the original memory unit operation), after the method ends the parameter value is changed. */public static void F1 (int[] a) {for (int i = 0; i < a.length; i++) {a[i]++;}} /** * After the parameter points to a new address in the method, the parameter will still point to the original address after the method ends. */public static void F2 (int[] a) {a = new int[]{2, 3, 4, 5, 6}}

Output:

2 3 4) 5 6
1 2 3) 4 5

Got it.

When Java passes a parameter, the difference between the value of the pass and the reference