On Tue, Mar 19, 2013 at 9:41 PM, AJ ONeal <coola...@gmail.com> wrote:
>
> This doesn't do too much to demystify anything, imo.
>>
>> > Now how that works... I have no idea, but I do believe that it's more
>> aesthetic and so that's how I write my code. I won't argue if you feel
>> differently.
>>
>
> I suppose I should have been more specific:
>
> This is an error foo():
>
> function () {} ()
>
> This is not (foo)()
>
> (function () {})()
>
> This is not (foo())
>
> (function () {}())
>
> So foo() is NOT identical to (foo)() where foo is function () {}.
>
Yes, they are.
> var foo = function() { return 1; }
> foo()
1
> (foo)()
1
> foo === (foo)
true
The parens around the function object reference are the grouping operator.
>
>
>
>>
>> It works because (foo)() is identical to foo().
>>
>
>
>> > The only reason that that even works at all is because there's magic
>> going on behind the scenes that hoists the anonymous function up to the top
>> (just like with variables) and then calls that function reference.
>>
>> What am I reading? Function expression are not hoisted.
>>
>
> var foo = function () {};
>
> foo gets hoisted
>
>
No, FunctionExpression (assignments) _do_ _not_ get hoisted.
> foo();
> var foo = function() {};
TypeError: undefined is not a function
>
>
> function foo() {};
>
> foo gets hoisted
>
Yes, because it's a Function Declaration
>
>
>
> for () {
> x = function () { console.log(i) };
> }
>
> What actually happens is this
>
> function something() {console.log};
> for () {
> x = something;
> }
>
No, the FunctionExpression assignment _does_ _not_ magically become a
FunctionDeclaration.
console.log( "x" in this );
for ( var i = 0; i < 1; i++ ) {
x = function () { console.log(i) }; // <--- NOT hoisted
}
false
console.log( "y" in this );
for ( var i = 0; i < 1; i++ ) {
function y() { console.log(i) }; // <--- hoisted
}
true
> I probably could word that better since that's not technically
> "hoisting", but in essence it's still hoisting. The creation of the
> function is pulled out of the loop in the same way `var x` would get pulled
> out and only the reference remains.
>
> Also setTimeout() takes optional parameters to pass to the callback:
>>
>> setTimeout(function(a) { console.log(a) }, 0, 'hello')
>> hello
>>
>
> Well, duh. Anyone with years of JavaScript experience knows that!
>
Are you sure? His point was that you were creating the IIFE in a loop
instead of passing the argument explicitly—which you can.
for ( var i = 0; i < 5; i++ ) {
setTimeout(function(a) { console.log(a); }, 0, i);
}
0
1
2
3
4
> However, years of wretched online tutorials do not. I agree that it's a
> good solution (and I should probably show that as an alternate way to do
> it. However, a common solution is to through in the IIFE.
>
>
>> Also there's Function.prototype.bind()
>>
>
> Also something I might include when I go back and edit.
>
Seems like there are a few things you need to edit.
Rick
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