> On 27 Apr 2017, at 09:54, Rick Mann <rm...@latencyzero.com> wrote: > >> >> On Apr 26, 2017, at 23:37 , Rien via swift-users <swift-users@swift.org> >> 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 surprised if the majority of ‘let’ constants never see any >> memory allocation at all. > > In my case, it's a field in a struct that's allocated elsewhere, and most > certainly exists, and is passed as a parameter, so it's 'let'. I really want > to be able to say "I promise I won't write to this", which seems reasonable > if the language is willing to give me an unsafe pointer in the first place.
IIRC this is the reason for the mutable and non-mutable version of the pointer. With the non-mutable pointer you cannot write to the pointee. (At least not without some additional code) But it is another step entirely to then allow a let assignment to be used as the pointee. I assume the Swift team decided to keep things simple and not allow let’s to be used as pointee’s. It could probably be done in theory, but I imagine that to be a nightmare case for compiler writers... Regards, Rien. > > How is the immutable version of withUnsafePointer different from the mutable > one? It passes the argument to the closure as a mutable pointer, which must > mean something different. > >> >> Regards, >> Rien >> >> Site: http://balancingrock.nl >> Blog: http://swiftrien.blogspot.com >> Github: http://github.com/Balancingrock >> Project: http://swiftfire.nl - A server for websites build in Swift >> >> >> >> >> >> >>> On 27 Apr 2017, at 08:31, Florent Bruneau via swift-users >>> <swift-users@swift.org> wrote: >>> >>> 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 to be a >>> var. >>> >>> There may be other subtleties I'm not aware of, though. >>> >>>> Le 27 avr. 2017 à 02:42, Rick Mann via swift-users <swift-users@swift.org> >>>> a écrit : >>>> >>>> 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 >>>> rm...@latencyzero.com >>>> >>>> >>>> _______________________________________________ >>>> swift-users mailing list >>>> swift-users@swift.org >>>> https://lists.swift.org/mailman/listinfo/swift-users >>> >>> -- >>> Florent Bruneau >>> >>> _______________________________________________ >>> swift-users mailing list >>> swift-users@swift.org >>> https://lists.swift.org/mailman/listinfo/swift-users >> >> _______________________________________________ >> swift-users mailing list >> swift-users@swift.org >> https://lists.swift.org/mailman/listinfo/swift-users > > > -- > Rick Mann > rm...@latencyzero.com _______________________________________________ swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
