I have:
var Init = (function() { my js goes here })(); And my JS executes correctly when the page is loaded. I also have:
$('form :checkbox').change(function() { Init(); }); But Firebug says Init is not a function.
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6 Answers
It isn't a function.
(function() { ... })() evaluates the anonymous function right then. And the result of the evaluation apparently does not return a function-object in this case :-)
Consider:
f = (function() { return "not a function :(" })() alert(f()) and
f = (function() { return function () { return "Yay!" } })() alert(f()) Happy coding :)
Here is a function which will "execute something once" and then "return that something to execute later". (See "You can either [assign] a function or call it; you can't do both..." from Slaks answer.) However, I wouldn't do it like this.
Init = (function () { function Init () { alert("whee!") } Init() return Init })() Init() Here is another solution (much shorter/cleaner) from CD Sanchez (see comment) which takes advantage of the fact that an assignment evaluates to the assigned value:
var Init; (Init = function Init () { alert ("wee"); })() 6 Comments
Cristian Sanchez
The last two snippets could simplified to:
var Init; (Init = function () { alert ("wee"); })()
Phillip Senn
I am now using var Init = (function() { return function() {...} })(); and it is working as I would like. Now whether that's the correct pattern or not, I don't know.
Zecc
Why not just write
function Init(){} Init(); ? It makes the intention clearer. |
In order for Init to execute as a function, your code within the self-executing function must return a function, and the only reason to do this is if you need to construct a specific function dynamically dependent upon some data states:
var Init = (function() { // some code return function () { // some dynamic code dependent upon your previous code }; }()); 
Comments
Init isn't a function; it's the result of calling the function.
You can either create a function or call it; you can't do both at once.
Technically, you could fix that by adding return arguments.callee; to return the function from the call.
However, that's a dumb idea.
You probably shouldn't be calling the function; you need to understand what you want your code to do.
7 Comments
dev-null-dweller
you can't do both at once - yes you can: (function Init(){ /*...*/ })();
SLaks
@dev: Wrong. That's a named expression, not a declaration. kangax.github.com/nfe/#named-expr
user113716
I think maybe "You can either assign a function or call it; you can't do both..." may get the point across better.
Cristian Sanchez
An assignment is a valid expression -- it returns the "value" of the object you assigned to the variable. With this in mind, you can enclose an assignment expression in parens and use the function call syntax to immediately call the function after assigning it:
var Init; (Init = function () { ... })(); Please note that (var Init = ...)() would not be valid because var statements are not expressions.user113716
@CD Sanchez: The assignment operator works right to left. The function will be invoked first, then its return value will be assigned to
Init. That's the issue in the question. The parentheses are not necessary in the code in your comment. |
Quick one Try replacing like this
var Init = function() { my js goes here }); and on load call Init
1 Comment
Phillip Senn
I'd like it to be a self-executing function, but also be able to call it later on.
you could do as above, but you could also do
function Init(){...}(); There's nothing to stop you from having a named self-executing function. If you want to avoid having a function named Init, you can do as CD Sanchez suggested and assign it in the execution.
The (); at the end makes it self executing. Wrapping the function in parentheses makes it anonymous. But it seems that you don't want it to be anonymous.
Comments
You may try declaring it this way:
(function Init(){ /*...*/ })(); But this will reduce usage of this function into it's body
Other way is to separate declaration from execution:
var Init = function(){ /*...*/ }, initResult = (function(){ return Init(); })(); 
7 Comments
SLaks
Wrong. That's a named expression, not a declaration. It's not visible outside of the function. kangax.github.com/nfe/#named-expr
Cristian Sanchez
By all means, correct me if I'm wrong, but I don't think
Init would defined in the enclosing scope using that. Perhaps you meant: var Init; (Init = function () { ... })();?user113716
Yeah, sorry but this isn't right. Aside from IE implementation bugs,
Init will not be available in the enclosing scope like OP wants.user113716
@CD Sanchez: That would actually have the same issue as the code in the question, unless you return a function from
Init.Cristian Sanchez
@patrick dw: I don't think so -- it would store the function in
Init and then execute the result of the assignment expression (which would be the function () {} part). I've used this in the past and it's worked perfectly. I'm not infallible though - if you see a flaw in my reasoning please let me know. |