How to assign values ​​between arrays in Java

The array name is the address of the *** element pointed to by the head pointer of the array. If you can understand it this way, then think about it with your thighs and you will know that a=b cannot realize the assignment of the array. Generally in C and C we A loop is used to assign a single value, similar to this:

1. for(int i=0;i<10;i )

2. a[i]=b[i];

way to achieve it, after learning C and C for so long, I use it more , and I don’t find it troublesome. But in JAVA, C is not the same as C. JAVA is very smart. It can use a=b to assign b to a. Pay attention to the assignment here. When using a=b in JAVA, in a When the content is output in the form of an array, it is exactly the same as the content of b, which shows that this method is feasible. This is how it is used in my program, but a bug results. At first I found that no matter whether I use a=b or a=a.clone(); the result is the same, which was a little confusing, but now I understand it completely. In order to illustrate this problem, let's demonstrate with a program:

private int[] subResources(int[] aa, int[] bb)  {     // 做减法     int []a=aa;     int []b=bb;     for (int i = 0; i < a.length; i++)      a[i] = a[i] - b[i];     return a;  }

This is used to implement a function that wants to subtract two arrays. Use the above two methods in the calling program

int []allo=pcb0.getAllocation().clone(); //方式1   //int []allo=pcb0.getAllocation()//方式2     int[] allocation = this.addResources(allo, request);

The different results brought about by passing in parameters are that when you modify the value of the passed in parameter in the function addResources, method 1 cannot change the original data, but method 2 can change the original data. When assigning value in method 1, it first creates a copy of the array, and then assigns the copy to the target array. In this way, the address of the copy array is certainly not the same as the address of the original array, so no matter how you change it, method 1 will not Will change the original data, but method 2 is different. Method 2 directly assigns the address of the data to the target data. In this way, the different array names of the two arrays actually refer to the same address, so of course the original value can be changed. .

Analyzed in this way, it is a bit like the pointer and reference value in function parameter passing in C. Since pointers are abandoned in Java, all pointing relationships use reference types. It takes a long time to use C It's easy to make such mistakes. It is not like passing value addResources (int *a,int *b) or addResources (int a[],int b[]) like in C. This will be a disadvantage. It seems that this is the only time in this place in my life. I have to focus more on learning. Now I understand why the basic questions at the job fair seem simple, but not everyone can answer them well.

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