of: The value to take the dynamic type of.
Returns: The dynamic type, which will be a value of metatype type.
Remark: If the parameter is statically of a protocol or protocol
composition type, the result will be an existential metatype
(P.Type for a protocol P), and will represent the type of the value
inside the existential container with the same protocol conformances
as the value. Otherwise, the result will be a concrete metatype
(T.Type for a non-protocol type T, or P.Protocol for a protocol
P). Normally, this will do what you mean, but one wart to be aware
of is when you use type(of:) in a generic context with a type
parameter bound to a protocol type:
func foo<T>(x: T) -> T.Type {
return type(of: x)
}
protocol P {}
func bar(x: P) {
foo(x: x) // Call foo with T == P
}
since the call to type(of:) inside foo only sees T as a concrete
type, foo will end up returning P.self instead of the dynamic type
inside x. This can be worked around by writing type(of: x as Any)
to get the dynamic type inside x as an Any.Type.
Returns the dynamic type of a value.
of: The value to take the dynamic type of. Returns: The dynamic type, which will be a value of metatype type.Remark: If the parameter is statically of a protocol or protocol composition type, the result will be an existential metatype (
P.Typefor a protocolP), and will represent the type of the value inside the existential container with the same protocol conformances as the value. Otherwise, the result will be a concrete metatype (T.Typefor a non-protocol typeT, orP.Protocolfor a protocolP). Normally, this will do what you mean, but one wart to be aware of is when you usetype(of:)in a generic context with a type parameter bound to a protocol type:func foo<T>(x: T) -> T.Type { return type(of: x) } protocol P {} func bar(x: P) { foo(x: x) // Call foo with T == P }since the call to
type(of:)insidefooonly seesTas a concrete type, foo will end up returningP.selfinstead of the dynamic type insidex. This can be worked around by writingtype(of: x as Any)to get the dynamic type insidexas anAny.Type.Declaration
functype<Type,Metatype>(of:Type) ->Metatype