CODEWITHSUNDEEP
javascriptinterpreter
AntonAL
AntonAL

Reputation: 17410

Order of functions in JavaScript

My question is based on the example from a book "Object Oriented JavaScript" (page 81 - Lexical Scope)

So, i understand from this example ...

function f1(){var a = 1; f2();}
function f2(){return a;}
f1();

... that:

a is not defined

But, how f1 get's to know about f2, which is defined after f1 ?

This behavior raises a question:

How JavaScript interpreter works ?

I assume, that it:

  1. scans the code and simply stores the functions, not assigned to any var, in a global environment
  2. Invokes a function in ad-hoc way: when there is no such a function in a global environment, then complain.

Upvotes: 7

Views: 4110

Answers (2)

T.J. Crowder
T.J. Crowder

Reputation: 1074525

Function declarations are processed upon entry into an executable context (e.g., the global context, or a function call), prior to any of the step-by-step code in the context being processed.

So in your code, these things happen (in this order):

  1. A "variable object" is created for the execution context.
  2. Entries (actually, literally, properties) on the "variable object" are created for every var and function declaration in the context (plus a few other things). In your case, that's f1 and f2. Initially the properties have the value undefined.
  3. All function declarations are processed, and so:
    • The f1 function is defined and assigned to its property on the variable object.
    • The f2 function is defined and assigned to its property on the variable object.
  4. The f1(); line is executed, calling the f1 function.
  5. The f1 code refers to f2, which it gets from the variable object, and so it's what we expect it to be (a reference to the f2 function).

The more interesting version is this:

f1();
function f1(){var a = 1; f2();}
function f2(){return a;}

...which happens in exactly the same order listed above, because both of the declarations are handled before the first line of step-by-step code.

Function declarations are different from function expressions, which just like any other expression are evaluated when they're reached in the step-by-step execution of the code. A function expression is any time you create a function and use it as a right-hand value, e.g., assign the result to a variable or pass it into another function. Like this:

var f2 = function() {
};

or this

setTimeout(function() {
    alert("Hi there");
}, 1000);

Note that we're using the result of the function statement as the right-hand value (in an assignment, or by passing it into a function). Those are not pre-processed upon entry into an execution context (e.g., not at Step 3 above), they're handled when the flow of code reaches them. Which leads to:

f1();
function f1(){var a = 1; f2();}
var f2 = function(){return a;};

...which fails, because f2 is undefined as of when it's called.

You can use a declared function's value as a right-hand value without turning it into a function expression (we do that all the time), so long as you do it in two separate statements. So:

alert("Beginning");
function foo() { ... }
setTimeout(foo, 100);

That happens in this order:

  1. foo is created (since it's defined by a declaration).
  2. The alert runs.
  3. The setTimeout runs.
  4. (Later) foo is called.

One last point: Although they should work, a function expression that includes a function name does not work reliably on all implementations and must, for now, be avoided:

var f = function foo() { ... }; // <== DON'T DO THIS

Or

setTimeout(function foo() {     // <== DON'T DO THIS
}, 1000);

Internet Explorer, in particular, has issues with those, and other implementations have at various times as well.

More to explore:

Upvotes: 16

fazo
fazo

Reputation: 1827

you don't have access to variable 'a' inside function f1, because function f2 is not defined inside f1 scope

if you define f2 inside f1:

function f1(){function f2(){return a;} var a = 1; f2();}
f1();

you don't have any problems

Upvotes: 0

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