What's happening behind the scene with implicitly unwrapped option when it's being evaluated?

[Boon]

What's happening behind the scene with implicitly unwrapped option?

Does the compile perform the unboxing of ImplicitlyUnwrappedOptional behind the scene, right before you use the value in an expression?  Since ImplicitlyUnwrappedOptional is a struct, can the same operation be done in Swift?

[Brent Royal-Gordon]

Implicitly unwrapped optionals are treated specially by the compiler; there's no way for us to write something similar.

(For that matter, regular optionals have some special behavior too; the automatic conversion of Foo to Optional<Foo> is unique to optionals. But they're far less magical than implicitly unwrapped optionals.)

On Wed, Jun 17, 2015 at 8:51 AM Boon <bo...@nanaimostudio.com> wrote:

What's happening behind the scene with implicitly unwrapped option?

Does the compile perform the unboxing of ImplicitlyUnwrappedOptional behind the scene, right before you use the value in an expression?  Since ImplicitlyUnwrappedOptional is a struct, can the same operation be done in Swift?

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