Programming Basics: Variable declaration, intitialiation, assignment, and scope

Question

How does scope interact with variable declaration, initialisation, and assignment? The definition of those terms based on what I have learned so far is listed below:

Declaration: States the type of a variable, and it's name/identifier. Variables must have been declared before they can be assigned or read.
Assignment: Throws away the existing value of a variable and replaces it with a new one, the old value is thrown away at the end of the assignment statement, so the value can be incremented or otherwise adjusted, for example: x = x + y;
Initialisation: The name used for the first assignment of a variable, before initialisation, a variable has a default value, in the case of objects, those objects have a null value. Initialisation can be done in conjunction with declaration.
Scope: The "lifespan" of a variable, a variable is in scope until the end of the code block, at which point the memory used to store that variable is freed up. In effect, the variable is deleted or "killed", when the code block ends.

What I don't know is how scope interacts with Declaration and assignment. While the scope of a variable seems to be based solely on the code block in which it is Declared, I don't know how assignment interacts with scope. For example:

public class exampleClass
{
   public static void main(String[] args) // using java for example
   {
      int x = 5; // x is declared here, and initialised with a value of 5
      for (int i = 0; i < 10; i++) // i is declared and initialised here
      {
         x = i; // x is assigned the value of i each iteration
      } // i goes out of scope here
   System.out.println(x); // the value of x is printed   
   } // x goes out of scope here
}

In this example, x is declared and initialised (do we just say initialised?) in the main method, and is in scope for that method. However, x is assigned a value in the while loop. What will be printed when this code is executed, but more importantly why? Will it print "5", or "9"?

I have seen code throw up compiler exceptions because of syntax that would imply that x should print 5. However when I run this example code, I get "9".

One final question, why is it that multiple variables can be declared and initialised inline:

int x = 1, y = 4, z = 6;

But variables cannot be assigned inline:

x = 1, y = 4, z = 6;

Essentially this question is more about the hows and why's than the results. I know that scope is based on where the variable was declared, I don't know why or how, and working with things I don't understand bothers me. In addition, while I do know that the example will give me "9", not "5", if I didn't already know, it'd be impossible to google it to find out, as an example of common knowledge this is rarely outright stated on websites like this one, thus, ironically making it less common knowledge. — Space Ostrich
Quite a few of these questions are likely to be language specific. There isn't a set of universal rules that apply to all languages. So I;'d suggest adding a language tag. — Damien_The_Unbeliever

deceze deceze · Accepted Answer · 2016-03-04T14:27:50

The distinction between declaration and initialisation can be blurry; some languages make a clear distinction between these actions, in others initialisation is declaration. If a variable is initialised while it's being declared, it doesn't matter which you call it.

However, x is assigned a value in the while loop. What will be printed when this code is executed, but more importantly why? Will it print "5", or "9"?

9, because that's the last value that has been assigned to it before you print it.

Scope: The "lifespan" of a variable, a variable is in scope until the end of the code block, at which point the memory used to store that variable is freed up. In effect, the variable is deleted or "killed", when the code block ends.

Yes and no. Scope defines in what parts of code a particular variable is available. Different languages can have very different definitions of their scoping rules. A variable is typically garbage collected (in languages where that's applicable) when it goes out of scope, when no piece of code has any further access to it. In a simple function block, that happens when the function ends.

However, see this Javascript example:

function foo() {
    var bar = 'baz';

    return function () {
        alert(bar);
    };
}

The inner function which is returned from this function still holds a reference to bar. Even if foo ends, bar is being closed over by a closure and is still in scope within the inner function. As long as a reference to that returned function exists, bar still exists.

Programming Basics: Variable declaration, intitialiation, assignment, and scope

1 Answers