jQuery: Rewriting Anonymous Callback to a Named Function

问题

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() { $('div:first').fadeOut(); }

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

h1 will slide up and div will fade out at the same time!

How can I make my second example work the same as the first one, where fadeOut() is called AFTER slideUp()?

回答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);

---

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.

---

Hope this example will help you understand:

function outer() {
  function inner() {};
  return inner;
}

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

来源：https://stackoverflow.com/questions/28268338/jquery-rewriting-anonymous-callback-to-a-named-function