[swift-users] Intended behavior or bug in Dictionary's new subscript(_:default:)?

Ben Cohen ben_cohen at apple.com
Sat Jun 17 23:12:19 CDT 2017

In order for this defaulting subscript technique to work as intended, the subscript { get } needs to be called, then the mutation happens, then the subscript { set } writes the mutated value back, including adding it for the first time it the default was needed.

Reference types, not being value types, skip the write-back part, because they shouldn’t need writing back – they should just get amended in place, because they’re reference types. 

Except this particular technique is relying on it.

This is probably worth a bug report, though I’m not sure if there’s an easy fix. The alternative is that Dictionary.subscript(_: default:) be made a mutating get that sets the default if not present, even without the set. There’s downsides to this though: you would no longer be able to use this defaulting subscript with immutable dictionaries, and getting a default value would add it which might be very unexpected.

> On Jun 16, 2017, at 1:40 PM, Jens Persson via swift-users <swift-users at swift.org> wrote:
> // Swift 4, Xcode 9 beta 1, default toolchain
> import Foundation
> var d1 = [Int : String]()
> d1[1, default: .init()].append("a")
> d1[2, default: .init()].append("b")
> d1[3, default: .init()].append("c")
> d1[1, default: .init()].append("d")
> print(d1) // [2: "b", 3: "c", 1: "ad"] as expected.
> var d2 = [Int : NSMutableString]()
> d2[1, default: .init()].append("a")
> d2[2, default: .init()].append("b")
> d2[3, default: .init()].append("c")
> d2[1, default: .init()].append("d")
> print(d2) // [:] but why?
> I know that NSMutableString is a reference type and String is a value type and that the default argument is an @autoclosure. I also know that the newly created NSMutableString instance is just released immediately after the append call, without being stored and retained in the Dictionary's storage.
> Is this the intended behavior and if so, please let me better understand how/why.
> /Jens
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