CODEWITHSUNDEEP
javascriptsetinterval
Afshin Mehrabani
Afshin Mehrabani

Reputation: 35009

setInterval with self-executing function

I want to run my function at the first time immediately (without timeout) so I do this:

setInterval(function(){
    alert("Boo");
}(), 1000);

The function executed at first time but in next intervals, nothing happened. Why?

Upvotes: 2

Views: 4225

Answers (7)

Brian Cray
Brian Cray

Reputation: 1275

Here's what I've done in the past as my approach:

(function loop () {
   // do stuff
   window.setTimeout(loop, 1000);
})();

Upvotes: 2

David Rettenbacher
David Rettenbacher

Reputation: 5120

Because what setInterval "sees" is the return of your self-executing function - undefined.

Try this instead:

    setInterval(function()
        {
             alert("Boo");
             return arguments.callee;

        }(), 1000);

Note:
As I Hate Lazy pointed out arguments.callee isn't permitted in Strict Mode.

Upvotes: 2

jAndy
jAndy

Reputation: 236202

The better question is, what are you actually trying to achieve ?

You don't return anything from the self-invoking function, so it will return the undefined value implicitly, which is passed over to setTimeout. After the initial call, the line would look like

setInterval(undefined, 1000);

which obviously, makes no sense and won't continue calling anything.

You either need to to return another function from your self-invoking function, or just don't use a self-invoking function and just pass over the function reference.

Update:

You updated your question. If you want to run a function "once" on init and after that use it for the interval, you could do something like

setInterval((function() {
    function loop() {
        alert('Boo');
    };

    loop();
    return loop;
}()), 1000);

Upvotes: 6

I Hate Lazy
I Hate Lazy

Reputation: 48819

It's because it is invoked, not passed. If you want an immediate invocation, just give it a name, and return it. 

setInterval(function foo() { 
    alert("boo"); 
    return foo; 
}(), 1000);

This is called a named function expression. You should be aware that in most browsers the foo identifier will only be available inside foo, but older IE has issues that causes the identifier to leak to the outer scope.

This is usually not an issue, but you should be aware of it when choosing the function name.

Upvotes: 3

Shmiddty
Shmiddty

Reputation: 13967

You could write it like this:

setInterval((function() {
    function boo() {
        alert("boo");
    };
    boo();
    return boo;
})(), 1000);​

But I'd suggest that you declare the function outside of the interval.

Upvotes: 1

Alexei Levenkov
Alexei Levenkov

Reputation: 100630

You self executing function does not return a function to use as argument of setInterval, so it shows alert before the call to setInterval and than setInterval is essentially no-op.

You can see Firefox complaining about empty call to setInterval if you want.

Upvotes: 1

jhummel
jhummel

Reputation: 1764

Set interval requires the function reference, not the undefined returned from the function. Just remove the '()' at the end of the anonymous function.

Upvotes: 3

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