No, you use checked instead of blah. If blah is type 'Blah?', Checked is type 'Blah'.

Edit to add: And unless something has changed since I last tried it, you can also just shadow blah so that inside that scope you're referring to the unwrapped one:

    if let blah = blah { /* use blah here, it's unwrapped */ }

It looks kind of silly, but I have used it where there was no better name for the unwrapped value.

this doesn't look better than c++ syntax.

I still don't understand the point of ! and ?

less is more, swift ought to know.

I don't think it's trying to be syntactically better than C++. It's an improvement in semantics. Once you unwrap an optional with "if let", it cannot cause an access violation.

You seem to be insisting on using ! and then blaming the language for letting you shoot your foot off with it...

Sorry, I don't get it.

the example you gave is like the following c++ code:

XXX *object = null;

if ((object = anotherpointer) != NULL)

{ ... }

The point is, ! is only useful, when it can detect null pointers during compiling time. but it doesn't.

the way ! is used, as suggested by your example, is also doable in C++, it is just a habit thing.

in c++, with good habit, you won't have problem. in Swift with bad habit, you will have the same problem.

then what good is !,

No, you don't seem to be getting what I'm saying. I'm saying don't use !, and showing you how to avoid using it in a common pattern. Using 'if let' instead of ! does give you compile time verification of correctness.

! is there for the small number of cases where that's not possible. And it makes it much more obvious where you've got a potential problem, because the '!' is a literal code smell. You can't really do that with c++.

Ever since the guard statement was introduced I think you should never, ever use the ! as operator. It still has some sense when using as a type, for example for IBOutlets.