.fadeIn() breaks when using .dequeue() rapidly

[rasamassen]

Because .fadeIn() ultimately doesn't mess with opacity, if you rapidly use the following, eventually the element's opacity is set to near zero by fadeIn()/fadeOut():

$('.foobar').hover(function() {

$('.sub').dequeue().stop(); $('.sub').delay(600).fadeTo('400',1);

}, function() {

$('.sub').dequeue().stop(); $('.sub').delay(600).fadeOut('400');

});

From what I understand in the documentation, fadeIn should bring an element to opacity:1. fadeTo can take to variable opacity in between.

Attached is an example that shows fadeIn failing to reach opacity:1 where fadeTo succeeds.

[rasamassen]

fadeIn test case

[rasamassen]

I double checked the definition of fadeIn: Display the matched elements by fading them to opaque. Thus, if opacity of an element called by fadeIn is set, it's opacity should be changed to 1 by the completion of fadeIn. This is not happening.

[rasamassen]

What may be a simple solution to this is to make fadeIn/fadeOut call fadeTo rather than directly calling animate. fadeTo has been developed to account for many more possibilities.

[rasamassen]

Working example of break at ​http://jsfiddle.net/K4FtG/ Break still present in 1.4.4rc2.

[rasamassen]

Still broken in 1.5

[Timmy Willison]

This is why we now have .stop(true, true). ​http://jsfiddle.net/timmywil/K4FtG/4/