Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

On 28 Apr 2017, at 09:10, Quinn The Eskimo! via swift-users wrote: > * If the memory is in a zone that uses fixed sized buckets, it may not be > possible to reduce the memory that you freed by reducing the size D’oh! For “to reduce the memory” read "to reuse the

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

On 28 Apr 2017, at 03:09, Rick Mann via swift-users wrote: > Because this runs on iOS, I'd like to reduce the amount of time I hang on to > memory as much as possible (there's a lot of parallel processing happening in > our pipeline). I'd also like to avoid copying the

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

You will need to take full control over allocation and deallocation. Here is a piece of code from my project SwifterSockets (on github, see the link in my signature) public func tipReceiverLoop( socket: Int32, bufferSize: Int, duration: TimeInterval, receiver: ReceiverProtocol?)

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

Yeah, okay. So: how do I do this in a way that is safe? -- Rick Mann rm...@latencyzero.com > On Apr 27, 2017, at 23:00, Rien wrote: > > To address your question: > > https://developer.apple.com/reference/foundation/data/1779823-withunsafemutablebytes > > "Warning >

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

To address your question: https://developer.apple.com/reference/foundation/data/1779823-withunsafemutablebytes "Warning The byte pointer argument should not be stored and used outside of the lifetime of the call to the closure." Which is exactly what you are doing, hence the code is unsafe (no

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

> On Apr 27, 2017, at 7:31 PM, Rick Mann wrote: > > >> On Apr 27, 2017, at 18:56 , Hooman Mehr wrote: >> >> You should be able to type your `dataBuffer ` as [Int8] (Byte array). Then >> you won’t need `withUnsafeMutableBytes`. You can simply call it

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

> On Apr 27, 2017, at 18:56 , Hooman Mehr wrote: > > You should be able to type your `dataBuffer ` as [Int8] (Byte array). Then > you won’t need `withUnsafeMutableBytes`. You can simply call it like this: > > self.request = c_library_call(, dataBuffer) // Call as if it is a C

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

So, the dataBuffer has a maximum initial size, then I resize it down to the actual resulting size (the data comes from a sensor and real-world vagaries change the resulting volume of data). Then it eventually gets passed to existing Objective-C++ code, and on to C++ code that wants a void* and

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

You should be able to type your `dataBuffer ` as [Int8] (Byte array). Then you won’t need `withUnsafeMutableBytes`. You can simply call it like this: self.request = c_library_call(, dataBuffer) // Call as if it is a C array It works because of C interoperability compiler magic. As long as the

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

> On Apr 27, 2017, at 01:48 , Alex Blewitt wrote: > ... > The let constant may not even be stored in a single place; if it's known to > be constant it can be in-lined at the point of use and potentially unpacked > and dead code elimination throw away the unused members, for

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

> On 27 Apr 2017, at 09:54, Rick Mann wrote: > >> >> On Apr 26, 2017, at 23:37 , Rien via swift-users >> wrote: >> >> 1) When you obtain a pointer, it can no longer be ensured by the compiler >> that you won’t write to it. >> 2) A ‘let’

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

> On Apr 26, 2017, at 23:37 , Rien via swift-users > wrote: > > 1) When you obtain a pointer, it can no longer be ensured by the compiler > that you won’t write to it. > 2) A ‘let’ variable (constant) allows way more optimizations than a ‘var’. I > would not be

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

1) When you obtain a pointer, it can no longer be ensured by the compiler that you won’t write to it. 2) A ‘let’ variable (constant) allows way more optimizations than a ‘var’. I would not be surprised if the majority of ‘let’ constants never see any memory allocation at all. Regards, Rien

Re: [swift-users] Why does withUnsafePointer(to:) require a var argument?

Hi Rick, My understanding on this is that withUnsafePointer() requires an inout argument because it has to take a reference to the variable in order to be able to derive its pointer. The languages requires inout arguments to be vars, leading to withUnsafePointer() requiring the passed object

[swift-users] Why does withUnsafePointer(to:) require a var argument?

We have withUnsafePointer(to:) and withUnsafeMutablePointer(to:). Why does the first take an inout parameter? The function names imply that the first will not modify the pointer (which I take to mean its contents), and it makes it quite clunky to pass in constant things. -- Rick Mann