Why can I manipulate a UIView's properties in Swift by passing it into a function if Swift passes by value not reference?

Question

Perhaps this is a silly question and I'm completely misunderstanding how passing variable in Swift works, but I was under the impression that Swift was pass by copy by default, and you could use the inout keyword to denote passing by reference.

How come then, if I have a function that has a UIView parameter, I can change, say, the backgroundColor of the UIView within that function, and it's reflected outside the function. If the function creates a local copy when passed, how is this affecting the passed variable?

Why does that not require inout?

Short: UIView is a reference type (as are all classes). – jtbandes Aug 23 '15 at 19:35 — jtbandes, Aug 23 '15 at 19:35

score 1 · Answer 1 · edited May 23 '17 at 12:06

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Because the UIView you pass around is a class instance: it is a pointer to begin with. So, in essence, you are passing that pointer by value, since you are not passing a reference to that pointer (like &pointer in C).

See Is Swift Pass By Value or Pass By Reference for an elaborate answer.

edited May 23 '17 at 12:06

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answered Aug 23 '15 at 22:28

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Why can I manipulate a UIView's properties in Swift by passing it into a function if Swift passes by value not reference?

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