How to use UnsafeMutablePointer in swift 4

Question

Objective-c has a concept of a pointer to a pointer. If you dereference the first pointer you can access the original

void makeFive(int *n) { *n = 5; } int n = 0; makeFive(&n); // n is now 5

When this is bridged to Swift 3 it becomes an UnsafeMutablePointer

func makeFive(_ n: UnsafeMutablePointer<Int>) { n.memory = 5 } var n: Int = 0 makeFive(&n) // n is now 5

However, as of Swift 4, this behavior has changed and the memory property is no longer available.

What would be the swift 4 equivalent of the makeFive(_:) function?

Update Thanks to Hamish, I now know that "memory" was renamed to pointee.

As of Swift 3, it's .pointee; but don't use UnsafeMutablePointer here. Use inout if you need to mutate a caller-side variable (and this shouldn't be that often). — Hamish
– Hamish, Commented Sep 20, 2017 at 18:52
Thanks for the correction and rare it certainly is! Otherwise I'd find it in stack overflow. I wasn't aware of the name change. I am mainly using UnsafeMutablePointer for objective-c compatibility since "inout" is a swift only feature. — DerrickHo328
– DerrickHo328, Commented Sep 20, 2017 at 19:01

Vini App · Accepted Answer · 2017-09-20 19:30:27Z

6

Please check : https://developer.apple.com/documentation/swift/unsafemutablepointer

func makeFive(_ n: UnsafeMutablePointer<Int>) { n.initialize(to: 5) } var n: Int = 0 makeFive(&n) // n is now 5

answered Sep 20, 2017 at 19:30

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Hamish Over a year ago

Be careful with this; the pointee is already initialised, so you shouldn't really be calling initialize(to:) on it. It will happen to work in this case because the Pointee is a trivial type (Int), but will produce undefined behaviour for non-trivial pointee types (I think at least; although the best case scenario is still leaking memory). The operation you want here is assignment to the pointee; and that's done by setting the pointee property of the pointer.

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