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Home/ Questions/Q 7197143
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T20:54:55+00:00 2026-05-28T20:54:55+00:00

I have taken the following example from Hands-on Node. It uses some reasonably advanced

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I have taken the following example from “Hands-on Node”. It uses some reasonably advanced techniques. I have copied the code verbatim and have tried to debug a few times but cannot figure out why nothing gets printed to my console.

var schedule = function(timeout, callbackfunction) {
    return {
        start: function() {
            setTimeout(callbackfunction, setTimeout)
        }
    }
};

(function() {
    var timeout = 1000;
    var count = 0;
    schedule(timeout, function doStuff() {
        console.log(++ count);
        schedule(timeout, doStuff);
    }).start(timeout);
})();
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  1. Editorial Team
    Editorial Team
    2026-05-28T20:54:56+00:00Added an answer on May 28, 2026 at 8:54 pm

    Aside from passing setTimeout instead of timeout (which will still allow it to run once), if this is meant to loop the timer, then there’s a flaw.

    The code never calls .start() again after the first time. You’d need to do this…

    (function() {
        var timeout = 1000;
        var count = 0;
        schedule(timeout, function doStuff() {
            console.log(++ count);
            schedule(timeout, doStuff).start(); // Invoke .start() each time
        }).start(); // Removed useless "timeout" argument
    })();
    

    I don’t know why someone would take this approach, as it seems overly complex.


    I’m not even sure why they were passing timeout to the inital .start(). That function doesn’t use any argument passed. I updated to remove it.


    If this was meant to teach about the benefit of closures, then this example really does little to do that.

    Yes, the .start() function does reference the timeout and callbackfunction parameters, but the object returned is used once and discarded, at which point we call schedule again, and pass it the same args.

    It would seem more useful as a demonstration if schedule just returned a function, and you retained a reference to that function. It would then only require the one call to schedule to hold the values.

    var schedule = function(timeout, callbackfunction) {
        return function() {
            setTimeout(callbackfunction, timeout)
        }
    };
    
    (function() {
        var count = 0;
        var fn;
        (fn = schedule(1000, function doStuff() { // 1. assign the function returned
            console.log(++ count);
            fn();  // 3. invoke the same function again
        }))();  // 2. invoke the returned function immediately
    })();
    
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