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
