Question

Please advice why primitives being used as method’s parameters do a copy of its value while objects are used as is?

Answer 1

In Java, all arguments are passed by value – but in the case of reference types (i.e. everything other than a primitive) the value of a variable isn’t the object itself – it’s a reference to the object. Thus that reference is copied into the method’s parameter, so it refers to the same object.

Note that this doesn’t just apply to method calls:

StringBuilder x = new StringBuilder();
StringBuilder y = x; // Copy the value of x, which is a *reference* 

y.append("Hello");
System.out.println(x); // Prints "Hello"

Here, x and y refer to the same object, even though they’re separate variables. Thus when the contents of that object is changed via the append call through the y variable, the change is visible via the x variable too.

I think of it as being a bit like giving someone the address of your house: if I give two people my home address, and one of them paints the door red, then when the second person visits the house, they’ll see the red door too. I’m not giving them my house itself, I’m giving them a way of getting to my house.

There are many, many articles about this – although unfortunately some will claim that objects are passed by reference in Java. They’re not – the references are passed by value, as I said above. Scott Stanchfield has a good article about this, amongst many others.