Java varargs如何在内存中实现

Question

From what I know, local vars and args are stored in the Stack Memory, which includes object references, whereas actual objects are stored in the heap memory. So what exactly happens when you use var-args ?

public static int[] test(int... x) {
    return x;
}

public static void main(String[] args) {
    int[] a = test(1,2,3,4,5);
    int[] b = test(6,7,8,9,0);
    System.out.println(a);
    System.out.println(b);
    for (int i : a) {
        System.out.println(i);
    }
    for (int i : b) {
        System.out.println(i);
    }

Here I believe that all the values passed as params to test in x are stored on the stack, hence when we call test(1,2,3,4,5) , we are using up stack space, correspondingly when we call test(6,7,8,9,0) we should be causing memory collision on the stack... but when I run the above, I get the following result.

[I@2db0f6b2
[I@3cd1f1c8
1
2
3
4
5
6
7
8
9
0

As can be seen, there is no corruption of items in a due to the 2nd call to test which generates b . Both seem to be stored differently.

Does this mean that the params are somehow stored on the heap? Would this mean that any call of the form function(param) translates to the value of param (primitive or memory reference) not necessarily lying on the stack memory?

Answer 1

Varags are just a syntactic sugaring on top of arrays - ie, using a int... parameter is the same as an int[] . Like all arrays, the array resides on the heap, and you pass a reference to it down on the stack.

Answer 2

x... is only a syntactic sugar for a x[] . Think its working procedure exactly same as raw array.

The reference itself could be stored on the heap if it is a member of a class or object, or on the stack if it is a local variable in a method. And primitive types can be stored on the heap if they are members of a class or object. - Source