dfri dfri - 5 months ago 14
Swift Question

Type erasure: do we risk non-reversibly loosing access to kept-alive data of the instance of the erased type, when erasing the type information?

Consider the following common simple type erasure scheme

protocol Foo {
associatedtype Bar
func bar() -> Bar

struct AnyFoo<Bar>: Foo {
private let _bar: () -> Bar

init<F: Foo>(_ foo: F) where F.Bar == Bar {
_bar = foo.bar
/* stores a reference to foo.bar,
so foo kept alive by ARC? */

func bar() -> Bar {
return _bar()

Assume the initializer argument
above is (intended to be) a temporary instance of a "large" type, from which we're only interested to slice out the information blueprinted by
(i.e., the

struct Huge { /* ... */ }

struct Foobar: Foo {
internal func bar() -> String {
return "foo"
let lotsOfData: Huge = Huge()

func getAnyFooStr() -> AnyFoo<String> {
let foobar = Foobar()
return AnyFoo(foobar)

let anyStrFoo = getAnyFooStr()
/* we can now access anyStrFoo.bar() (-> "foo") from our
erased type, but do lotsOfData of the underlying seemingly
temporary Foobar() instance still "live", unreachable? */

  • Q: Will we still keep the remaining content of
    alive and unreachable in memory due to the fact that closures are reference types? And if so, I assume we would never be able to reclaim access to this lost-but-alive content?

(I tried the above with
as a class, monitoring the (lack of a)
call, but I've confused myself so much the last hour that I need some non-self verification for this, especially for the case when
is a value type)

Xcode 8.0 / Swift 3.


Yes, and it has nothing to do with type erasure :) The closure Foobar().bar keeps the instance that it is bound to (with all its properties) alive as long as the closure is alive.

Here is a simplified example:

class Huge {
    deinit { print("deinit Huge") }

struct Foobar {
    internal func bar() -> String {
        return "foo"
    let lotsOfData: Huge = Huge()

do {
    let fb  = Foobar().bar // the type of `fb` is `() -> String`
    print("still alive ...")
print("... out of scope now")


still alive ...
deinit Huge
... out of scope now