Variable hoisting

Question

alert(myVar1);
return false;
var myVar1;

Above code throws error in IE, FF and Opera stating that return statement has to come in the function. But it works (shows undefined) in Safari and Chrome.

The above code has been written in global scope. Outside all the functions.

Any reason?

Answer 1

In javaScript Variables are moved to the top of script and then run. So when you run it will do

var myVar1;
alert(myVar1);
return false;

This is because javascript doesnt really have a true sense of lexical scoping. This is why its considered best practise to have all your variables declared at the top of the area they will be used to prevent hoisting causing a problem. JSLint will moan about this.

This is a good article that explains it http://www.adequatelygood.com/2010/2/JavaScript-Scoping-and-Hoisting

The return is invalid. If you want to do a true hoisting example (taken from the link above) do

var foo = 1; 
function bar() { 
    if (!foo) { 
        var foo = 10; 
    } 
    alert(foo); 
} 
bar();

This will alert 10

EDIT AFTER COMMENT

Below is my understanding and I have read it somewhere but can't find all the sources that I read so am open to correction.

This Alerts thanks to the differences in the JavaScript JIT. TraceMonkey(http://ejohn.org/blog/tracemonkey/) I believe will take the JavaScript and do a quick static analysis and then do JIT and then try run it. If that fails then obviously nothing works.

V8 doesnt do the static analysis and moves to the JIT then runs so something. Its more akin to python. If you run the script in the Developer console(ctrl+shift+j in windows) in Chrome it will throw an error but also run to give you the alert.

Answer 2

Sometimes hoisting is explained in a way which may give wrong impression, i.e. the variables and functions are hoisted by JavaScript engine as if they were physically moved on top, which is not actually right, as demonstrated by the code below:

console.log(a);
var a = 'Hello World!';

What we see on console is undefined, not 'Hello World', so we got the behavior of the following code

var a;
console.log(a);
a = 'Hello World!';

not the behavior of

var a = 'Hello World!';
console.log(a);

which you may get the impression from the variables and functions declaration being moved to top statement.

But JavaScript is not actually moving your code anywhere. You need to understand execution context in JavaScript. It has two phases creation phase and execution phase. In creation phase memory space is created for these variables and functions, and people seem to confuse this step with hoisting. JavaScript is actually not moving your code anywhere, what happens is JavaScript has created memory space for all your code i.e. variables and functions, functions can be placed fully in memory but in case of variables the assignments are processed in execution phase of the execution context. So when you do var a = 'Hello World!', JavaScript engine knows the value of a when it starts executing it in execution phase of execution context, so it puts a placeholder undefined, and all variables are initially set to undefined in JavaScript. So it is not good to rely on hoisting and see undefined. So it is always good to declare variables and functions on top of your code.

Answer 3

Section 12.9 (page 75) of ECMA-262 edition 3 states:

An ECMAScript program is considered syntactically incorrect if it contains a return statement that is not within a FunctionBody.

That is, a return outside of a function is a syntax error. If a syntax error occurs, no code is run. Think about your example as if you had written:

alert(myVar1);
return false;
syntax error))))))))))))))))));

In addition, section 16 (page 157) states:

An implementation may treat any instance of the following kinds of runtime errors as a syntax error and therefore report it early:

• Improper uses of return, break, and continue.

Firefox's engine et. al. (i.e. those JavaScript implementations which allow return in the global scope) may be conforming, assuming the following clause (in the same section) allows for implementation definition of return in the global scope:

An implementation shall report all errors as specified, except for the following:

• An implementation may provide additional types, values, objects, properties, and functions beyond those described in this specification. This may cause constructs (such as looking up a variable in the global scope) to have implementation-defined behaviour instead of throwing an error (such as ReferenceError).

Answer 4

This code makes little sense:

• The var myVar1 will never be ran.
• The return false; will not return a thing since you're not in a function

Opera, IE and FF are right to throw an error because this code is really not valid since you're not able to return unless you are in a function.

If it works in Safari and Chrome it must be because the javascript engine they use is ready to handle buggy code. My guess would be that they see the "return" and drop or replace it with some kind of break.

More information about functions: http://www.w3schools.com/js/js_functions.asp

Answer 5

It is JavaScript Hoisting thing, Simply in other words, we are trying print the value of the variable , which is not holding any value

Javascript will render the above code as:-

var myVar1
alert (myVar1)
return false

To clarify more i refered the link javascript hoisting: http://www.ufthelp.com/2014/11/JavaScript-Hoisting.html