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  1. PO
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    <p>using modulus this is pretty easy:</p> <pre><code>var array = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], index = 8, n = 3; var result = []; for (var i = index - n, len = array.length; i &lt;= index + n; i++) { result.push(array[(i + len) % len]); } console.log(result); </code></pre> <p>Edit JSFiddle: <a href="http://jsfiddle.net/cmymS/" rel="nofollow">http://jsfiddle.net/cmymS/</a></p> <p>An explanation for the modulo operator in this example:</p> <p>The modulo operator handles going over the boundaries. Inside of the for loop, <code>i</code> is a 'virtual' index in the array that can be less than 0 or <code>&gt;= len</code>. You can think of this as having infinite repetitions of your array both on the left and right of your actual array. If <code>i &lt; 0</code>, we are accessing one of the virtual arrays to the left, if <code>i &gt;= len</code>, we're accessing one of the virtual arrays to the right. The index transformation <code>(i + len) % len</code> then handles translating 'virtual' indices back into actual array indices.</p> <p>There are two boundary conditions and the case of accessing a 'normal' index:</p> <p>a) <code>i &lt; 0</code>: Consider e.g. <code>i = -3</code>.</p> <p><code>(i + len)</code> will shift the position by one whole array so that we're working in the virtual array to the right, but pointing at the same element. The modulus has no effect since <code>len - 3</code> is smaller than <code>len</code>.</p> <p>b) <code>i &gt;= len</code>: Consider e.g. <code>i = len + 4</code></p> <p><code>(i + len)</code> will shift the position one array to the right. In our example, <code>(i + len)</code> will be <code>(len + 4 + len)</code> but the modulo neutralizes those shifts so we get <code>4</code>.</p> <p>c) <code>i</code> is a valid index in the original array <code>(i + len)</code> will shift the position one array to the right, but this is reset by the modulus since <code>0 &lt;= i &lt; len</code>. <code>(i + len) % len = i</code>.</p>
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    1. USStefan Seemayer
    1. USStefan Seemayer
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    1. COVery nice. How does the modulo work in this context? I've used it before for stuff like if(i % 2 == 0) for alternating class names and the like.
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    2. COUpdated with my attempt at an explanation.
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      1. USStefan Seemayer
    3. COExplanation greatly appreciated. I've moved yours to the accepted answer mostly due to the fact that VisioN's answer causes undefined's to appear when you hit an overflow as you mentioned in the comment, although both answers are great examples of approaching the problem.
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