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Unable to run application twice, even after quitting first one #3063
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The second delay should probably be > 2sec as it takes more than a 0 time for the quit to execute and completely shut down an app instance. |
Thanks Andy, I updated the example code above to 4 seconds and the issue still exists. |
I thought it would, but good to check! |
This likely relates to #920 - but this issue is a clear description that applies to more OS. |
Possibly related to #639? |
Hi Andy, although this was almost a year ago, was the root cause of this ever found? Any way to restart without completely stopping the binary? |
I think you can follow the chain of issues to a PR that was opened but never completed. |
Describe the bug:
An app cannot be stopped and either started again or replaced with a new, working instance of an app.
When starting an app and showing a window, quitting it, and trying to start the app again, no window is shown the second time. The app cannot be run twice even when the first instance of it is removed. Code provided below shows this in a minimally reproducible example.
Looks like the code at L69 of the loop blocks on the second run and doesn't go any further with debugging. Andrew from slack reckons "It's possible we don't signal
funcQueue
".Suspected related issue.
To Reproduce:
Expectation from the following code is that a window should appear repeatedly, but it is only shown once. The same issue happens even with
app.New()
placed within the loop.Device (please complete the following information):
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