When debugging Swift code, can I get a typed reference to an object given just its address?

Question

When debugging Objective-C code in LLDB, I often create variables that refer to objects in memory using just their address. For example:

(lldb) po self.view
<UIView: 0x7ff5f7a18430; frame = (0 64; 320 504); autoresize = W+H; layer = <CALayer: 0x7ff5f7a192e0>>
(lldb) e CALayer* $layer = (CALayer*) 0x7ff5f7a192e0
(lldb) e $layer.borderWidth
(CGFloat) 17ドル = 0

Given just an object's type and its address in memory, I'm able to inspect and manipulate it.

Is this impossible when debugging Swift code?

Answer 1

(lldb) e let $layer = unsafeBitCast(0x7fd120f474b0, CALayer.self)

• Bill: unsafeBitCast is even cooler. How did you find out about these APIs? None of them seem to be in the docs.
• Darren: @Bill I found them by poking around the Swift standard library header (Command+Click on a standard library function or symbol) . All of the interesting functions and classes have "Unsafe" in their name or comments.
• sfaxon: I posted the updated lldb expression that got me started on Xcode 7.3 here: stackoverflow.com/questions/29441418/…
• Martin R: In Swift 3: e let $layer = unsafeBitCast(0x7fd120f474b0, to: CALayer.self)