How are Int and String accepted as AnyHashable?

Question

How come I can do this?

    var dict = [AnyHashable : Int]()
    dict[NSObject()] = 1
    dict[\"\"] = 2

This implies that NSObject

Answer 1

Consider that Optional is an enum, which is also a value type – and yet you're freely able to convert a String to an Optional. The answer is simply that the compiler implicitly performs these conversions for you.

If we look at the SIL emitted for the following code:

let i: AnyHashable = 5

We can see that the compiler inserts a call to _swift_convertToAnyHashable:

  // allocate memory to store i, and get the address.
  alloc_global @main.i : Swift.AnyHashable, loc "main.swift":9:5, scope 1 // id: %2
  %3 = global_addr @main.i : Swift.AnyHashable : $*AnyHashable, loc "main.swift":9:5, scope 1 // user: %9

  // allocate temporary storage for the Int, and intialise it to 5.
  %4 = alloc_stack $Int, loc "main.swift":9:22, scope 1 // users: %7, %10, %9
  %5 = integer_literal $Builtin.Int64, 5, loc "main.swift":9:22, scope 1 // user: %6
  %6 = struct $Int (%5 : $Builtin.Int64), loc "main.swift":9:22, scope 1 // user: %7
  store %6 to %4 : $*Int, loc "main.swift":9:22, scope 1 // id: %7

  // call _swift_convertToAnyHashable, passing in the address of i to store the result, and the address of the temporary storage for the Int.
  // function_ref _swift_convertToAnyHashable
  %8 = function_ref @_swift_convertToAnyHashable : $@convention(thin) <τ_0_0 where τ_0_0 : Hashable> (@in τ_0_0) -> @out AnyHashable, loc "main.swift":9:22, scope 1 // user: %9
  %9 = apply %8(%3, %4) : $@convention(thin) <τ_0_0 where τ_0_0 : Hashable> (@in τ_0_0) -> @out AnyHashable, loc "main.swift":9:22, scope 1

  // deallocate temporary storage.
  dealloc_stack %4 : $*Int, loc "main.swift":9:22, scope 1 // id: %10

Looking in AnyHashable.swift, we can see the function with the silgen name of _swift_convertToAnyHashable, which simply invokes AnyHashable's initialiser.

@_silgen_name("_swift_convertToAnyHashable")
public // COMPILER_INTRINSIC
func _convertToAnyHashable(_ value: H) -> AnyHashable {
  return AnyHashable(value)
}

Therefore the above code is just equivalent to:

let i = AnyHashable(5)

---

Although it's curious that the standard library also implements an extension for Dictionary (which @OOPer shows), allowing for a dictionary with a Key of type AnyHashable to be subscripted by any _Hashable conforming type (I don't believe there are any types that conform to _Hashable, but not Hashable).

The subscript itself should work fine without a special overload for _Hashable keys. Instead the default subscript (which would take an AnyHashable key) could just be used with the above implicit conversion, as the following example shows:

struct Foo {
    subscript(hashable: AnyHashable) -> Any {
        return hashable.base
    }
}

let f = Foo()
print(f["yo"]) // yo

Edit: In Swift 4, both the aforementioned subscript overload and _Hashable have been removed from the stdlib by this commit with the description:

We have an implicit conversion to AnyHashable, so there's no need to have the special subscript on Dictionary at all.

Which confirms my suspicion.