refactor: make debugger nullable

[bluwy]

Description

Often times we do:

isDebug && debug('...')

const start = isDebug ? performance.now() : 0

Where isDebug is process.env.DEBUG, however that doesn't always mean the debug will run if DEBUG="blabla" and we're logging with the vite:time namespace.

To simplify this, the PR refactors as debug?.('...') instead.

Additional context

Might make debug logs slightly more accurate.

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What is the purpose of this pull request?

• Bug fix
• New Feature
• Documentation update
• Other

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[sapphi-red]

LGTM

Maybe returning Debugger | undefined from createDebugger makes it more simple?
Then, we can write debug?.('something') instead of debug.enabled && debug('something').

[bluwy]

Not sure why I didn't think of that, that feels simpler too 🤔

[bluwy]

Updated. I like this new pattern instead, it makes all debug logs lazy by default now.

[patak-dev]

+1 for the new pattern 💚