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copied!<p>Geoff has outlined one approach (using <code>setInterval</code> and clearing it from the callback), so I'll outline the other: Using <code>setTimeout</code> from the callback. Something like</p> <pre><code>m = 10 do drawCallback = -> draw m, 150 m += 10 setTimeout drawCallback, 1000 unless m > 100 </code></pre> <p>Note that there is a subtle timing difference between the two approaches that you should be aware of: <code>setInterval func, 1000</code> will run the function once every 1000ms; the chained <code>setTimeout</code> will put a 1000ms delay between each function call. So if <code>draw</code> took 100ms, say, the chained <code>setTimeout</code> would be equivalent to <code>setInterval func, 1100</code>. It probably doesn't matter, but it's worth being aware of.</p> <p><strong>Bonus approach</strong>: You don't <em>have</em> to abandon your loop; you could just set all the timeouts from it at once:</p> <pre><code>for m in [10..100] by 10 do (m) -> setTimeout (-> draw(m, 150)), 100 * m </code></pre> <p>The <code>do (m)</code> is necessary so that the closure passed to <code>setTimeout</code> sees each value of <code>m</code>, not just its final value in the loop. See my article <a href="http://pragprog.com/magazines/2011-05/a-coffeescript-intervention" rel="noreferrer">A CoffeeScript Intervention</a> for more info on this.</p> <p><strong>Finally</strong>: I know this all seems very confusing at first, but timing in JS is actually very simple because the language is single-threaded. That means that events you schedule with <code>setTimeout</code> or <code>setInterval</code> or any other async function will never occur <em>during</em> a loop, even if the loop is infinite. They only occur after all of your code has finished executing. I talk about this in a little more detail in <a href="http://pragprog.com/book/tbcoffee/coffeescript" rel="noreferrer">my book on CoffeeScript</a>.</p>

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