Jquery: Rewriting Anonymous Callback To A Named Function

If I do this: $('h1').slideUp('slow', function() { $('div:first').fadeOut(); }); h1 will slide up, then the first div will fade out. However, if I do this: function last() { $('di

Solution 1:

You don't need to use the function return value (which you get by calling the function), but the function body:

$('h1').slideUp('slow', last);

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What you did is the same as this:

var returned = last();             // call to last returns undefined// so returned has the value undefined
$('h1').slideUp('slow', returned); // simply sending undefined as a callback

So you were just executing the last function inline, and then passing the return value (which is undefined since it returns nothing) as a parameter to the slideUp's callback function.

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Hope this example will help you understand:

functionouter() {
  functioninner() {};
  return inner;
}

alert(outer);    // returns the outer function bodyalert(outer());  // returns the outer function's return value, which is the inner function