Passing Arrays by Value and by Reference

These are example from a c# book that I am reading just having a little trouble grasping what this example is actually doing would like an explanation to help me further understand what is happening here.

        //creates and initialzes firstArray
        int[] firstArray = { 1, 2, 3 };

        //Copy the reference in variable firstArray and assign it to firstarraycopy
        int[] firstArrayCopy = firstArray;

        Console.WriteLine("Test passing firstArray reference by value");


        Console.Write("\nContents of firstArray " +
            "Before calling FirstDouble:\n\t");

        //display contents of firstArray with forloop using counter
        for (int i = 0; i < firstArray.Length; i++)
            Console.Write("{0} ", firstArray[i]);

        //pass variable firstArray by value to FirstDouble
        FirstDouble(firstArray);

        Console.Write("\n\nContents of firstArray after " +
            "calling FirstDouble\n\t");

        //display contents of firstArray
        for (int i = 0; i < firstArray.Length; i++)
            Console.Write("{0} ", firstArray[i]); 

        // test whether reference was changed by FirstDouble
        if (firstArray == firstArrayCopy)
            Console.WriteLine(
                "\n\nThe references refer to the same array");
        else
            Console.WriteLine(
                "\n\nThe references refer to different arrays");

       //method firstdouble with a parameter array
       public static void FirstDouble(int[] array)
    {
        //double each elements value
        for (int i = 0; i < array.Length; i++)
            array[i] *= 2;

        //create new object and assign its reference to array
        array = new int[] { 11, 12, 13 };

Basically there is the code what I would like to know is that the book is saying if the array is passed by value than the original caller does not get modified by the method(from what i understand). So towards the end of method FirstDouble they try and assign local variable array to a new set of elements which fails and the new values of the original caller when displayed are 2,4,6.

Now my confusion is how did the for loop in method FirstDouble modify the original caller firstArray to 2,4,6 if it was passed by value. I thought the value should remain 1,2,3.

Thanks in advance

Solution 1:

The key to understanding this is to know the difference between a value type and a reference type.

For example, consider a typical value type, int.

int a = 1;
int b = a;
a++;

After this code has executed, a has the value 2, and b has the value 1. Because int is a value type, b = a takes a copy of the value of a.

Now consider a class:

MyClass a = new MyClass();
a.MyProperty = 1;
MyClass b = a;
a.MyProperty = 2;

Because classes are reference types, b = a merely assigns the reference rather than the value. So b and a both refer to the same object. Hence, after a.MyProperty = 2 executes, b.MyProperty == 2 since a and b refer to the same object.

Considering the code in your question, an array is a reference type and so for this function:

public static void FirstDouble(int[] array)

the variable array is actually a reference, because int[] is a reference type. So array is a reference that is passed by value.

Thus, modifications made to array inside the function are actually applied to the int[] object to which array refers. And so those modifications are visible to all references that refer to that same object. And that includes the reference that the caller holds.

Now, if we look at the implementation of this function:

public static void FirstDouble(int[] array)
{
    //double each elements value
    for (int i = 0; i < array.Length; i++)
        array[i] *= 2;

    //create new object and assign its reference to array
    array = new int[] { 11, 12, 13 };
}

there is one further complication. The for loop simply doubles each element of the int[] that is passed to the function. That's the modification that the caller sees. The second part is the assignment of a new int[] object to the local variable array. This is not visible to the caller because all it does is to change the target of the reference array. And since the reference array is passed by value, the caller does not see that new object.

If the function had been declared like this:

public static void FirstDouble(ref int[] array)

then the reference array would have been passed by reference and the caller would see the newly created object { 11, 12, 13 } when the function returned.

Solution 2:

What a confusing use of terms!

To clarify,

1. for a method foo(int[] myArray), "passing a reference (object) by value" actually means "passing a copy of the object's address (reference)". The value of this 'copy', ie. myArray, is initially the Address (reference) of the original object, meaning it points to the original object. Hence, any change to the content pointed to by myArray will affect the content of the original object.

   However, since the 'value' of myArray itself is a copy, any change to this 'value' will not affect the original object nor its contents.

2. for a method foo(ref int[] refArray), "passing a reference (object) by reference" means "passing the object's address (reference) itself (not a copy)". That means refArray is actually the original address of the object itself, not a copy. Hence, any change to the 'value' of refArray, or the content pointed to by refArray is a direct change on the original object itself.

Solution 3:

All method parameters are passed by value unless you specifically see ref or out.

Arrays are reference types. This means that you're passing a reference by value.

The reference itself is only changed when you assign a new array to it, which is why those assignments aren't reflected in the caller. When you de-reference the object (the array here) and modify the underlying value you aren't changing the variable, just what it points to. This change will be "seen" by the caller as well, even though the variable (i.e. what it points to) remains constant.