A question about one line of code in use-on-unmount.ts

[mefengl]

@RobinMalfait When I read the source code of use-on-unmount.ts:

headlessui/packages/@headlessui-react/src/hooks/use-on-unmount.ts

Lines 10 to 20 in 6a88fd5

	trulyUnmounted.current = false
	return () => {
	trulyUnmounted.current = true
	microTask(() => {
	if (!trulyUnmounted.current) return
	stableCb()
	})
	}
	}, [stableCb])

It was really difficult for me to understand why there is a need for if (!trulyUnmounted.current) return. The value has already been updated before.

So I tried asking ChatGPT. It mentioned that this ensures the value will not be changed elsewhere, but it couldn't provide a reasonable example. It would be greatly appreciated if you could explain the reason or provide an example of why this line is necessary.

Alternatively, if this is just a coding pattern, it would also be helpful to know.

[mefengl]

Got an answer from LLM, wonder if it's right:

In the useOnUnmount function, the unmount callback function returned by useEffect is executed when the component is unmounted. However, in some cases, the component may be rerendered after unmounting, such as when server-side rendering or using React.lazy to load components. In this case, if the passed callback function is executed directly without any checks, it may cause errors or unexpected behavior.
Therefore, the line if (!trulyUnmounted.current) return is added in the callback function. It checks if trulyUnmounted.current is true. If it is, it means the component has been truly unmounted, and the callback function can be executed. If it is not, it means the component may be rerendering, and the callback function should not be executed, so it returns undefined without performing any operation. This way, the passed callback function can be executed only when it’s truly necessary after the component is unmounted, avoiding unnecessary errors and unexpected behavior.