What part of an array local variable is stored on the stack? And which part is on the heap?

Question

Suppose we have the following code:

void method() {
    int[] test = new int[3];
    test[0] = 0;
    test[1] = 1;
    test[2] = 2;
}

From what I read from Jon Skeet's post on SO, the new int[3] part is equivalent to:

public class ArrayInt3 {
    public readonly int length = 3;
    public int value0;
    public int value1;
    public int value2;
}

So does that mean test (a reference to ArrayInt3) is on the stack? And does that mean ArrayInt3 is on the heap? And I suppose value0, value1, and value2 are on the heap as well (i.e. 0, 1, 2 in this example)?

So in total, there is 4 objects on the heap, correct?

Sotirios Delimanolis · Accepted Answer · 2014-09-24 15:57:23Z

1

There will be no object on the stack. On the stack, there will be a reference value to the single int[] object stored in the heap.

You have to start making the distinction between objects, variables, and values.

Local variables are part of the method stackframe, which is on the stack. So the reference value to the int[] will be stored in the variable, on the stack.

edited Sep 24, 2014 at 15:57

answered Sep 24, 2014 at 15:52

Sotirios Delimanolis

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8 Comments

Sotirios Delimanolis Over a year ago

@SomeRandomGuy Those are not objects. First, ArrayInt3 is a type. The rest are values. They are part of the single int[] object. They are not independent.

Sotirios Delimanolis Over a year ago

@SomeRandomGuy If they were initialized, they would be 4 reference values pointing to 4 other objects on the heap.

Sotirios Delimanolis Over a year ago

@SomeRandomGuy No, there is a single object on the heap, the int[] object.

SomeRandomGuy Over a year ago

Assuming value0, value1, and value2 are initialized, then what do we have on the heap exactly? value0, value1, value2, and ArrayInt3, right? I apologize for asking this poorly.

Sotirios Delimanolis Over a year ago

@SomeRandomGuy You'll have memory allocated for an instance of type ArrayInt3. That memory will be large enough to fit exactly those fields and their value.

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