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copied!<p>Declaring the variable outside the function makes the function use the same object every time.</p> <p>An example (with an integer instead of an object, for simplicity's sake):</p> <pre><code>var c = { theMethod: function () { var m1 = 0; return function (theParameter) { m1++; console.log( m1 ); } }() }; c.theMethod(); c.theMethod(); // output: 1 2 var d = { theMethod: function () { return function (theParameter) { var m1 = 0; m1++; console.log( m1 ); } }() }; d.theMethod(); d.theMethod(); // output: 1 1 </code></pre> <p>The self-invoking function works like this:</p> <pre><code>var c = { theMethod: function () { var m1 = 0; return function (theParameter) { m1++; console.log( m1 ); } }() }; </code></pre> <p>When the object <code>c</code> is created, the self-invoking function invokes itself and <code>theMethod</code> now equals the return value of that function. In this case the return value is another function.</p> <pre><code>c.theMethod = function( theParameter ) { m1++; console.log( m1 ); }; </code></pre> <p>The variable <code>m1</code> is available to the function because it was in scope when the function was defined.</p> <p>From now on when you call <code>c.theMethod()</code> you are always executing the inner function that was returned from the self-invoking function, which itself executed only once at the time the object was declared.</p> <p>The self-invoking function works just like any function. Consider:</p> <pre><code>var c = { theMethod: parseInt( someVariable, 10 ) }; </code></pre> <p>You don't expect <code>parseInt()</code> to execute every time you use the <code>c.theMethod</code> variable. Replace <code>parseInt</code> with an anonymous function as in the original and it's exactly the same thing.</p>

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