Analyze the following code:
class Test {
public static void main(String[] args) {
int[] x = {1, 2, 3, 4};
int[] y = x;
x = new int[2];
for (int i = 0; i < y.length; i++) {
System.out.print(y[i] + " ");
}
}
}
The answer for the following code is A. But why the answer is not B ?
Let's assume that {1, 2, 3, 4} has memory address M.
When assigning x to {1, 2, 3, 4}, you're assigning a reference to the memory address of {1, 2, 3, 4}, i.e. x will point to M.
When assigning y = x, then y will refer M.
After that, you're changing the reference where x points to, let's say it's N.
So, when printing, y points to M (which is the address of {1, 2, 3, 4}), but x holds reference to N (which is new int[2]). And here comes the difference.
To give you a more clear explanation:
int[] x = {1, 2, 3, 4}; // step 1
int[] y = x; // step 2
x = new int[2]; // step 3
In the third step, when the x changes, y is not affected because you are changing the reference of x, not the address of the array. So it doesn't affect the value of y.
Because int[] y = x; the array y will have some variables like this: {1,2,3,4}.
This line x = new int[2]; will creating a different space for x pointing to it.
So the result will be {1,2,3,4} of the array y
The answer is definitely A. I will explain it a little bit more step by step:
int[] x = {1, 2, 3, 4};
int[] y = x;
From the above lines, we now know that x and y are equal to each other in values and their length is 4.
x = new int[2];
We have now changed the value of x, but we haven't changed the value of y.
So now x has a length of 2 which is {0, 0}, and y still has a length of 4 which is {1, 2, 3, 4}, and here comes the difference...
for (int i = 0; i < y.length; i++)
In the above line, we already know that y.length = 4, so it's i < 4.
System.out.print(y[i] + " ");
The above line will now print the array of values of y respectively which are:
1
2
3
4
xandyare references.