int[]b=a : why does a change if i change b?

Question

int[]a={1,3,4,5};
int[]b=a;
b[2]=10;
System.out.println(a[2]);

the result is 10 !
can someone explain to me why a is touched?
is b a reference to a?

Answer 1

Good question. :>

Try doing:

int a[] = {1,2,3,4};
System.out.println(a);

You will realise that a is actually a memory address. When you are assigning a to b, you are assigning address of a to b and hence, if either values in a or b are altered, the changes will reflect on both sides.

This act is called the shallow-copy of a to b.
On the other hand, a deep-copy would be to assign the values of the array to another array by using a loop.

for (int i = 0; i < array_size; i++)
{
b[i] = a[i];
}

In this case, the memory address of both arrays are different and changes made to an array will not reflect on the other.

Answer 2

I dont know if this right or not but this is how understand the code in my head.

int[]a={1,3,4,5}; //declaring array 'a'
int[]b=a; // changing array a to 'int[]b={1,3,4,5};'
b[2]=10; // stating that the third number in array b is now 10 instead of 4.
System.out.println(a[2]); // array a= array b. Thus the fourth number in the array is the same.

thus, im thinking they (a and b) are 2 names for the same array. I might be wrong though, i just enjoyed trying to figure it out lol

Answer 3

An array declaration is mainly a pointer to a memory address –it doesn't hold actual values.