Using private API

Question

Created 1w

Replies 2

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A question for Apple engineers. Why is it that using a private API (for example, finding the corner radius of a device's screen) can result in the app being rejected for review or even an account being blocked, while using a private compiler API (for example, @_spi) has no consequences?

Answered by DTS Engineer in 859493022

Private APIs have not been officially released for public use. As such, they are not documented and engineering is not keeping track of them or maintaining them in system software updates. As such, they are subject to change at any time without notice. If you build an app using Private APIs your app may cease functioning, crash, or start operating in unexpected ways at any time.

Disallowing Private API usage in apps available on the App Store is protecting both you and your customers from unexpected inconveniences.

If your app requires functionality that is not provided by public APIs, please consider filing an enhancement request asking for the APIs that you require.

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Answer 1

Claude31 OP

1w

I'm not Apple engineer, but my understanding is that private API may change without notice.

Hence, if you use in your app, what worked in an OS version may crash with the next release.

That's bad for user and for reputation of iOS / MacOS.

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Answer 2

DTS Engineer OP

Apple

1w

Recommended

Private APIs have not been officially released for public use. As such, they are not documented and engineering is not keeping track of them or maintaining them in system software updates. As such, they are subject to change at any time without notice. If you build an app using Private APIs your app may cease functioning, crash, or start operating in unexpected ways at any time.

Disallowing Private API usage in apps available on the App Store is protecting both you and your customers from unexpected inconveniences.

If your app requires functionality that is not provided by public APIs, please consider filing an enhancement request asking for the APIs that you require.

0