[swift-evolution] [swift-users] UnsafeMutablePointer<String> vs. UnsafeMutablePointer<Int>

[Adrian Zubarev]

So basically if my application terminates without errors or crashes I can be sure that the OS will free any memory it used.

Now I feel safe using UnsafeMutablePointers in swift :)



-- 
Adrian Zubarev
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Am 1. Juni 2016 um 18:19:01, Austin Zheng (austinzheng at gmail.com) schrieb:

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

Best,
Austin

On Jun 1, 2016, at 9:13 AM, Adrian Zubarev via swift-users <swift-users at 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_rose at apple.com) schrieb:


On May 21, 2016, at 01:48, Adrian Zubarev via swift-users <swift-users at 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


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