CODEWITHSUNDEEP
swift
Boon
Boon

Reputation: 41480

When will a variable be inferred as an implicitly unwrapped optional by Swift compiler?

When will a variable be inferred as an implicitly unwrapped optional by Swift compiler? Or the variable has to always be declared with ! for it to be treated as such?

Upvotes: 2

Views: 218

Answers (1)

Jiaaro
Jiaaro

Reputation: 76918

The type inference occurs at compile time. The only way for a variable to be inferred as an implicitly unwrapped optional is to assign one to it (either directly, or by assigning the return value of a function, which returns an implicitly unwrapped optional).

Basically, the "optional-ness" of a variable is part of it's type. The type of the following variables, x and y is implicitly unwrapped Optional Int (for both), and most importantly, the type system does not consider this the "same type" as Int. (Though you can use it interchangeably with an Int as long as it is not nil)

let x: Int! = 7

fund make_y() -> Int! {
  return 7
}

let y = make_y()

Upvotes: 1

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