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 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
