Ios 如何解除锁定从Swift结构引用的UnsafemeutablePointer
如果我有这样一个Swift结构:Ios 如何解除锁定从Swift结构引用的UnsafemeutablePointer,ios,swift,cocoa-touch,swift2,Ios,Swift,Cocoa Touch,Swift2,如果我有这样一个Swift结构: struct ViewBox { let pointer: UnsafeMutablePointer<UIView> init() { pointer = UnsafeMutablePointer<UIView>.alloc(1) } } struct ViewBox { class WrappedPointer() { let pointer: UnsafeMut
struct ViewBox {
let pointer: UnsafeMutablePointer<UIView>
init() {
pointer = UnsafeMutablePointer<UIView>.alloc(1)
}
}
struct ViewBox {
class WrappedPointer() {
let pointer: UnsafeMutablePointer<UIView>
init() {
pointer = UnsafeMutablePointer<UIView>.alloc(1)
}
deinit {
pointer.dealloc(1)
}
}
let wrappedPointer = WrappedPointer()
}
struct视图框{
let指针:非女性化指针
init(){
指针=非女性化指针.alloc(1)
}
}
当结构被解除分配时,我应该如何确保指针被正确地解除分配?我不能对Swift结构使用deinit
或dealloc
方法
或者我不必在意,它会自动发生 可以将指针包装在类中。大概是这样的:
struct ViewBox {
let pointer: UnsafeMutablePointer<UIView>
init() {
pointer = UnsafeMutablePointer<UIView>.alloc(1)
}
}
struct ViewBox {
class WrappedPointer() {
let pointer: UnsafeMutablePointer<UIView>
init() {
pointer = UnsafeMutablePointer<UIView>.alloc(1)
}
deinit {
pointer.dealloc(1)
}
}
let wrappedPointer = WrappedPointer()
}
struct视图框{
类WrappedPointer(){
let指针:非女性化指针
init(){
指针=非女性化指针.alloc(1)
}
脱硝{
指针.解除锁定(1)
}
}
设wrappedPointer=wrappedPointer()
}
有趣的想法。所以你认为没有直接的方法可以做到这一点/不,我不这么认为。