How to delay a class from executing?

Question

So I have this:

(function () {
   // work
}).defer(1000);

For my admittedly weak understanding of javascript this executes an anonymous class after a second of wait time.

So I wanted to change it to something like this:

var newClass = function() {
   // work
};
newClass().defer(1000);

But this doesn't give me the delay for some reason. I feel like these two code samples are exactly the same, I am wrong in this assumption?

Edit:

I'm using this defer method, sorry I thought it was a javascript thing.

"Classes", i.e. constructor functions, are normal functions. — Felix Kling, Jun 12 '13 at 15:55
*" I feel like these two code samples are exactly the same, I am wrong in this assumption?"* Yes, in the first example you are not executing the function you defined. In the second case you are. What you want is `newClass.defer(1000)`. — Felix Kling, Jun 12 '13 at 15:57

score 3 · Answer 1 · answered Jun 12 '13 at 15:49

3

newClass().defer

This calls the function, then calls defer() on whatever the function returns.

answered Jun 12 '13 at 15:49

SLaks

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Is there any way I can call a sleep/wait for a second in javascript? – Grammin Jun 12 '13 at 15:55
@Grammin: [What do I do if I want a JavaScript version of sleep()?](http://stackoverflow.com/q/951021) – Felix Kling Jun 12 '13 at 15:56
@Grammin: No; you need to use asynchrony. (`setTimeout` or promises) – SLaks Jun 12 '13 at 17:50

score 2 · Accepted Answer · answered Jun 12 '13 at 15:50

2

you don't define a defer method, so here is one that works (almost) like you expect:

Function.prototype.defer=function(delay){ return setTimeout(this, delay || 100); };
var newClass = function() {
   alert ("working");
   // work
};
newClass.defer(1000);

answered Jun 12 '13 at 15:50

dandavis

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I've updated my answer to reflect what defer method I'm using. – Grammin Jun 12 '13 at 15:54
ok, then stike my first line, the invocation syntax is the same. Just don't call the function before you defer() it. – dandavis Jun 12 '13 at 15:55

How to delay a class from executing?

2 Answers2