Re: [swift-users] UnsafeMutablePointer vs. UnsafeMutablePointer

[Austin Zheng via swift-users]

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 
>  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
>  
> init(integer: Int) {
> self.pointer = UnsafeMutablePointer.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 
> ) schrieb:
> 
>> 
>>> On May 21, 2016, at 01:48, Adrian Zubarev via swift-users 
>>> > wrote:
>>> 
>>> I played around with UnsafeMutablePointer and realized one behavior:
>>> 
>>> let pString = UnsafeMutablePointer.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.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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Re: [swift-users] UnsafeMutablePointer vs. UnsafeMutablePointer

[Adrian Zubarev via swift-users]

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
 
init(integer: Int) {
self.pointer = UnsafeMutablePointer.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) schrieb:


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

I played around with UnsafeMutablePointer and realized one behavior:

let pString = UnsafeMutablePointer.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.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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Re: [swift-users] UnsafeMutablePointer vs. UnsafeMutablePointer

[Dmitri Gribenko via swift-users]

On Sat, May 21, 2016 at 2:15 AM, Adrian Zubarev via swift-users
 wrote:
> So basically if I do something like this I should be on the safe side:

Yes, this code is safe.  If you just want to store a contiguous buffer
of elements of the same type, you should consider using Array.  It has
methods that will allow you to operate on the unsafe pointer to the
memory if you need that for speed, but it will do the memory
management for you.

> Does the UnsafeMutablePointer reserves me a safe portion of memory (by safe
> I mean which isn’t used by anyone at the time I will start using it)?

Yes, you will become a unique owner of the newly allocated chunk of
memory.  (Same as malloc() in C.)

Dmitri

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Re: [swift-users] UnsafeMutablePointer vs. UnsafeMutablePointer

[Adrian Zubarev via swift-users]

So basically if I do something like this I should be on the safe side:

public class Characters {
 
private let reference: UnsafeMutablePointer

var characters: [Character] {
 
var characters = [Character]()
 
for index in 0...alloc(self.count)
self.reference.initializeFrom(characters)
}
 
deinit {
 
self.reference.destroy(self.count)
self.reference.dealloc(self.count)
}
}
Or do I have to fix something?

Here I don’t walk out of the boundary I allocate.

Does the UnsafeMutablePointer reserves me a safe portion of memory (by safe I 
mean which isn’t used by anyone at the time I will start using it)?

Sure another pointer could be created and hack into my memory but that wasn’t 
the question. :)

Thanks.



-- 
Adrian Zubarev
Sent with Airmail

Am 21. Mai 2016 bei 11:04:10, Dmitri Gribenko (griboz...@gmail.com) schrieb:

Hi Adrian,  

On Sat, May 21, 2016 at 1:48 AM, Adrian Zubarev via swift-users  
 wrote:  
> I played around with UnsafeMutablePointer and realized one behavior:  
>  
> let pString = UnsafeMutablePointer.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.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?  

Yes, you did.  

In the String case the crash is not guaranteed (it is a segmentation  
fault, not a controlled trap). Someone else's valid String can happen  
to be located immediately before your memory chunk, and then the code  
would "just work", loading that other string.  

Dereferencing (reading or writing into) an out-of-bounds pointer is  
undefined behavior in Swift.  

Dmitri  

--  
main(i,j){for(i=2;;i++){for(j=2;j*/  
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