[api-minor] Sort PopupAnnotations already on the worker-thread (PR 11535 follow-up) #15283
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…535 follow-up) By doing this in the worker-thread this code will only need to run *once*, whereas currently re-rendering of a page forces this to be repeated (e.g. after it's been scrolled out-of-view and then back into view again).
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/botio test |
From: Bot.io (Linux m4)ReceivedCommand cmd_test from @Snuffleupagus received. Current queue size: 0 Live output at: http://54.241.84.105:8877/90980c0f93a2643/output.txt |
From: Bot.io (Windows)ReceivedCommand cmd_test from @Snuffleupagus received. Current queue size: 0 Live output at: http://54.193.163.58:8877/4e9ff5acf599168/output.txt |
| @@ -656,7 +656,32 @@ class Page { | |||
| } | |||
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| return Promise.all(annotationPromises).then(function (annotations) { | |||
| return annotations.filter(annotation => !!annotation); | |||
| if (annotations.length === 0) { | |||
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I think it can be simplified in using sort function.
I just tried:
["a",1, 2, "d", "e", 3, "aa"].sort((a, b) => typeof a === typeof b ? 0 : (typeof a === "string" ? -1 : +1))and it works.
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To be honest, I'm not finding such a code-snippet particularly easy to read/understand at a glance.
Also, wouldn't you still need to filter out any empty entries first? If so, that'd effectively require two loops through the Array rather than just the one.
From: Bot.io (Linux m4)FailedFull output at http://54.241.84.105:8877/90980c0f93a2643/output.txt Total script time: 25.89 mins
Image differences available at: http://54.241.84.105:8877/90980c0f93a2643/reftest-analyzer.html#web=eq.log |
From: Bot.io (Windows)FailedFull output at http://54.193.163.58:8877/4e9ff5acf599168/output.txt Total script time: 30.22 mins
Image differences available at: http://54.193.163.58:8877/4e9ff5acf599168/reftest-analyzer.html#web=eq.log |
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Looks good! |
By doing this in the worker-thread this code will only need to run once, whereas currently re-rendering of a page forces this to be repeated (e.g. after it's been scrolled out-of-view and then back into view again).