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    <p>In JavaScript, you have function <em>statements</em> and function <em>expressions</em>. The first declare named functions, the latter evaluate to a named or anonymous function. You're using a function expression.</p> <p>What I think you want to do is a <em>call</em> expression. All you have to do is to immediately call your function that returns <code>i</code>.</p> <pre><code>function f() { var i, array = []; for (i = 0; i &lt; 3; i++) { // The extra parentheses around the function are unnecessary here. // But this is more idiomatic, as it shares syntax with how function // expressions are introduced in statements. // I.e. you could just copy-paste it anywhere. array[i] = (function () { return i; })(); // don't rely on automatic semicolon insertion } return array; } </code></pre> <p>Also note that, in your problematic example, all closures return <code>3</code> because all of them capture the same variable <code>i</code>.</p> <p>EDIT: To make the previous paragraph more clear, if you really wanted to have 3 distinct closures, you'd have to create a new scope for each one. Unlike other languages, Javascript doesn't create scope just by opening a block. It only creates scope in functions. Here are two ways of doing this:</p> <pre><code>function f() { var i, array = []; for (i = 0; i &lt; 3; i++) { // With a parameter in an inner function array[i] = (function (n) { return function () { // A new n is captured every time return n; }; })(i); } return array; } function f() { var i, array = []; for (i = 0; i &lt; 3; i++) { array[i] = (function () { // With a variable in an inner function var n = i; return function () { // A new n is captured every time return n; }; })(); } return array; } </code></pre> <p>The next example, however, is <strong>wrong</strong>, because even though <code>n</code> is declared in the <code>for</code> block, it will be as if it has been declared at the top of the function and only initialized in the <code>for</code> block. Remember, this is JavaScript, not Java, not C, not C#, not &lt;whatever bracketed language&gt;:</p> <pre><code>function f() { var i, array = []; for (i = 0; i &lt; 3; i++) { var n = i; array[i] = function () { return n; }; } return array; } </code></pre> <p>Equivalent code:</p> <pre><code>function f() { var i, array = []; var n; for (i = 0; i &lt; 3; i++) { n = i; array[i] = function () { return n; }; } return array; } </code></pre>
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    1. POClosure confusion
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    1. CODoing `array[i] = (function() { return i; })()` is the same as doing `array[i] = i;`, so you might as well get rid of the function.
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    2. COthank you Paulo for the reply.it worked! that clears the confusion.
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    3. COAnd in the original question, `array[i]` is being assigned a function, not just `i`. You're assigning it an IIFE, which means its contents will be used, which is just `i`.
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