Why is my function call that should be scheduled by setTimeout executed immediately? [duplicate]

Answer 1

This line here is the problem:

setTimeout(doRequest(url,proxys[proxy]), proxytimeout);

Writing doRequest() is actually calling the function. What you want is to pass the function itself:

setTimeout(doRequest, proxytime, url, proxys[proxy]);

Answer 2

As you give the function to the setTimeout in that form, the function is executed instead of passed to the setTimeout. You have three alternatives to make it work:

Give first the function, then the timeout and the parameters as the last arguments:

setTimeout(doRequest, proxytimeout, url, proxys[proxy]);

Or just write a string that will be evaluated:

setTimeout('doRequest('+url+','+proxys[proxy]+')', proxytimeout);

Third style is to pass an anonymous function that calls the function. Note that in this case, you have to do it in a closure to prevent the values from changing in the loop, so it gets a bit tricky:

(function(u, p, t) {
    setTimeout(function() { doRequest(u, p); }, t);
})(url, proxys[proxy], proxytimeout);

The second format is a bit hacky, but works nevertheless if the arguments are scalar values (strings, ints etc). The third format is a bit unclear, so in this case the first option will obviously work best for you.

Answer 3

You're misunderstanding the setTimeout function.

The setTimeout function takes a function and executes it later.
By writing setTimeout(doRequest(url,proxys[proxy]), proxytimeout), you're _calling the doRequest function (immediately), and passing the result (assuming that it returns another function) to setTimeout.

You need to pass doRequest's parameters to setTimeout, like this:

setTimeout(doRequest, proxytimeout, url, proxys[proxy]);

This will pass setTimeout the doRequest function itself (without calling it first), and will also pass it the parameters to give to doRequest when it finally calls it.