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    <p>The good news is that JavaScript is single threaded; this means that solutions will generally work well with "shared" variables, i.e. no mutex locks are required.</p> <p>If you want to serialize asynch tasks, followed by a completion callback you could use this helper function:</p> <pre><code>function serializeTasks(arr, fn, done) { var current = 0; fn(function iterate() { if (++current &lt; arr.length) { fn(iterate, arr[current]); } else { done(); } }, arr[current]); } </code></pre> <p>The first argument is the array of values that needs to be passed in each pass, the second argument is a loop callback (explained below) and the last argument is the completion callback function.</p> <p>This is the loop callback function:</p> <pre><code>function loopFn(nextTask, value) { myFunc1(value, nextTask); } </code></pre> <p>The first argument that's passed is a function that will execute the next task, it's meant to be passed to your asynch function. The second argument is the current entry of your array of values.</p> <p>Let's assume the asynch task looks like this:</p> <pre><code>function myFunc1(value, callback) { console.log(value); callback(); } </code></pre> <p>It prints the value and afterwards it invokes the callback; simple.</p> <p>Then, to set the whole thing in motion:</p> <pre><code>serializeTasks([1,2, 3], loopFn, function() { console.log('done'); }); </code></pre> <p><a href="http://jsbin.com/elebok/1/"><strong>Demo</strong></a></p> <p>To parallelize them, you need a different function:</p> <pre><code>function parallelizeTasks(arr, fn, done) { var total = arr.length, doneTask = function() { if (--total === 0) { done(); } }; arr.forEach(function(value) { fn(doneTask, value); }); } </code></pre> <p>And your loop function will be this (only parameter name changes):</p> <pre><code>function loopFn(doneTask, value) { myFunc1(value, doneTask); } </code></pre> <p><a href="http://jsbin.com/elebok/2/"><strong>Demo</strong></a></p>
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    1. POHow to synch JavaScript callbacks?
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    1. COThx @Jack, I'll need to go through this in detail later, have to sleep now :)
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    2. CO@Xavier_Ex I've added a parallel version too :)
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    3. COThanks for the effort and thank everybody for making suggestions. All posts are great but only this post answered my 2nd question. Most of the suggested solutions involved the use of a counter of some sort, but the promise sounds promising. I think I'll explore that a bit more :)
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