This shouldn't be something you need to worry about. The mechanism the OS uses to handle memory per process is different from the mechanism your process uses to allocate memory, and the OS should reclaim all the memory that your app used (whether it was 'leaked' or not).
More info: http://stackoverflow.com/questions/2975831/is-leaked-memory-freed-up-when-the-program-exits <http://stackoverflow.com/questions/2975831/is-leaked-memory-freed-up-when-the-program-exits> Best, Austin > On Jun 1, 2016, at 9:13 AM, Adrian Zubarev via swift-users > <swift-users@swift.org> wrote: > > I’ve got one more question that bothers me. > > Lets say I’ve got a class that might look something like this: > > class Reference { > > var pointer: UnsafeMutablePointer<Int> > > init(integer: Int) { > self.pointer = UnsafeMutablePointer<Int>.alloc(1) > self.pointer.initialize(integer) > } > > deinit { > self.pointer.destroy(1) > self.pointer.dealloc(1) > } > } > Let talk about ARC here. If I use optionals here and release manually the > reference deinit will be called and we’re happy here: > > var reference: Reference? = Reference(integer: 123456789) > reference = nil > If I don’t use optionals because I want my value to exist while the > application is running, deinit will never be called but my application > terminates just fine (SIGKILL?): > > let reference = Reference(integer: 123456789) > Doesn’t this create a memory leak? > > How do I solve this problem, especially if don’t know whats inside the > Reference type (assume I’m a different developer who only has access to the > framework but not its implementation)? > > > > > -- > Adrian Zubarev > Sent with Airmail > > Am 23. Mai 2016 um 18:31:43, Jordan Rose (jordan_r...@apple.com > <mailto:jordan_r...@apple.com>) schrieb: > >> >>> On May 21, 2016, at 01:48, Adrian Zubarev via swift-users >>> <swift-users@swift.org <mailto:swift-users@swift.org>> wrote: >>> >>> I played around with UnsafeMutablePointer and realized one behavior: >>> >>> let pString = UnsafeMutablePointer<String>.alloc(1) >>> pString.initialize("test") >>> pString.predecessor().memory // will crash ax expected >>> pString.predecessor() == pString.advancedBy(-1) // true >>> pString.destroy() >>> pString.dealloc(1) >>> >>> where >>> >>> let iInt = UnsafeMutablePointer<String>.alloc(1) >>> iInt.initialize("test") >>> iInt.predecessor().memory // will not crash >>> iInt.predecessor() == iInt.advancedBy(-1) // true >>> iInt.predecessor().memory = 42 // did I just modified some memory I don't >>> own? >>> iInt.destroy() >>> iInt.dealloc(1) >>> >>> Is this intended? This is really the case where its unsafe. >>> >> >> Dmitri’s answers are all better for this specific discussion, but in >> general, “unsafe” in Swift means “if you don’t follow the rules, this may >> crash, may silently corrupt memory or do other bad things, may cause other >> code to be optimized out or miscompiled, may be harmless”. In this >> particular case, it’d be hard to check for the validity of the pointer while >> also being fast and binary-compatible with C. >> >> Jordan >> > > > _______________________________________________ > swift-users mailing list > swift-users@swift.org > https://lists.swift.org/mailman/listinfo/swift-users
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