CODEWITHSUNDEEP
iosobjective-cxcodeswift
Aggressor
Aggressor

Reputation: 13551

How Does Parameter Type (void *) In Objective-C Translate To UnsafePointer<()> In Swift?

This is an example of an Objective-C function I needed to translate to Swift

- (void)image:(UIImage *)image didFinishSavingWithError:(NSError *)error contextInfo: (void *) contextInfo;

After much googling I finally found an answer where I could convert the (void *) to a valid Swift type. This type is apparently expressed as UnsafePointer<()>.

Do you know why (void*) translates to UnsafePointer<()> in Swift?

What is the syntax <()> called and what does it mean?

Upvotes: 2

Views: 915

Answers (1)

matt
matt

Reputation: 535222

<...> is a generic specifier. It resolves the generic placeholder of the generic type to whose name it is appended.

For example, Array is a generic, where the placeholder is its element type. So Array<String> is the type of an array whose generic placeholder is specified as being a String - meaning, an Array whose elements are strings. You may say [String], but Array<String> is equally valid.

Similarly, UnsafePointer is a generic, where the placeholder is the type of thing it points to. So UnsafePointer<Float> is an unsafe pointer to a Float - which might be the first float in a C array of Floats. (That actually comes up in real-life Swift programming.)

() is an empty tuple type, also known as Void. For example, a function that returns no value returns () (or Void).

So, putting it all together, UnsafePointer<()> (or UnsafePointer<Void>) is an unsafe pointer-to-void - which, by golly, is exactly what void* is.

Upvotes: 4

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