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Effects: sync .stop/.finish with .tick to maintain proper queue (#3497) #3501

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Effects: sync .stop/.finish with .tick to maintain proper queue (#3497) #3501

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ae094b1
added a list that maintains removed animations by .finish and .stop. …
preethi26 Jan 16, 2017
c6753d3
followed coding guidelines
preethi26 Jan 16, 2017
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18 changes: 15 additions & 3 deletions src/effects.js
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Original file line number Diff line number Diff line change
Expand Up @@ -483,7 +483,7 @@ jQuery.speed = function( speed, easing, fn ) {

return opt;
};

var lis = [];
jQuery.fn.extend( {
fadeTo: function( speed, to, easing, callback ) {

Expand Down Expand Up @@ -552,6 +552,7 @@ jQuery.fn.extend( {

timers[ index ].anim.stop( gotoEnd );
dequeue = false;
lis.push( index );
timers.splice( index, 1 );
}
}
Expand Down Expand Up @@ -590,6 +591,7 @@ jQuery.fn.extend( {
for ( index = timers.length; index--; ) {
if ( timers[ index ].elem === this && timers[ index ].queue === type ) {
timers[ index ].anim.stop( true );
lis.push( index );
timers.splice( index, 1 );
}
}
Expand Down Expand Up @@ -631,15 +633,25 @@ jQuery.each( {
} );

jQuery.timers = [];
var oldLength = 0;
jQuery.fx.tick = function() {
var timer,
i = 0,
timers = jQuery.timers;

fxNow = jQuery.now();
oldLength = timers.length;
for ( ; i < oldLength; i++ ) {
var removedTimers = 0;

for ( ; i < timers.length; i++ ) {
timer = timers[ i ];
for ( var j = 0; j < lis.length; j++ ) {
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I'm hoping to avoid O(N²) operations and extra global variables, and was thinking we'd do this by checking/clearing timer.elem since that's already cleared in an always callback.

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@preethi26 preethi26 Jan 16, 2017 •
edited

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ok ! not a problem.
I would like to know if the second bug about extra processing was handled?
As I wasn't sure about it I did not mark it in the checklist.

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Probably not, but we'll see. This PR will also require unit tests that don't pass without the updated code.

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@preethi26 preethi26 Jan 30, 2017 •
edited

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hey @gibson042 ,

jQuery.timers = [];
jQuery.fx.tick = function() {

var timer,
	i = 0,
	timers = jQuery.timers;

fxNow = jQuery.now();
for ( ; i < timers.length; i++ ) {
	timer = timers[ i ];
	var x  = timers.indexOf(timer)
	
	if (  !timer()  ) {
		if( x > timers.indexOf(timer)) {
			i = i - ( x - timers.indexOf(timer) )
		}

		if( timers[ i ] === timer ) {
			timers.splice( i--, 1 );
		}
	}
}

if ( !timers.length ) {
	jQuery.fx.stop();
}
fxNow = undefined;

};
This is the edited [ jQuery.fx.tick()] https://github.com/jquery/jquery/blob/master/src/effects.js#L644) function.
I would like to know if this fixes the first part of the bug before proceeding further.

Thank you,
Preethi.

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@preethi26 preethi26 Feb 3, 2017 •
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@gibson042. Sorry, just realized that it wouldn't work. Coz we need to check the value of "timer" 's index in between consecutive iterations ( which is not done here ).

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@gibson042
error_bug_1

In every iteration, making the value of the iterator equal to the value of the previous timer that successfully ran + 1.

if ( lis[ j ] <= i ) {
removedTimers++;
}
lis.splice( j, 1 );
}
i = i - removedTimers;
timer = timers[ i ];

// Checks the timer has not already been removed
if ( !timer() && timers[ i ] === timer ) {
Expand Down