CODEWITHSUNDEEP
iosswiftcocoa-touch
user89862
user89862

Reputation:

Calling removeFromSuperview() from deinit

The following code doesn't seem to work for me.

class SomeSubView: UIView {

    deinit {
        removeFromSuperview()
    }

}


let view = UIView()

var subview: SomeSubView! = SomeSubView()
var subview2: SomeSubView! = SomeSubView()

view.addSubview(subview)
view.addSubview(subview2)

subview = nil
subview = SomeSubView()
view.addSubview(subview)

print(view.subviews) //contains three views, should contain two

I also had a print() inside the deinit, so I'm pretty sure it is called. I could call subview.removeFromSuperview() before making the expression subview = nil, that works fine.

I'm only curious why this limitation exists, what kind of stuff can not be done inside deinit? I feel what I'm trying to do should work ...

Upvotes: 1

Views: 1159

Answers (1)

rob mayoff
rob mayoff

Reputation: 385590

It is never necessary to call removeFromSuperview() on self in deninit. A superview retains its subviews, so it's impossible for a view to be deallocated while it is a subview.

Setting subview = nil does not deallocate the object referenced by subview. It simply stops making subview reference that object.

Upvotes: 3

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