Swift Switch access nested properties within 'case'

Question

Following this code snippet. I'm trying to understand if it's possible to access nested properties of the object within Switch statement, without the need to unwrap properties inside the 'case' itself (avoid unneeded closures). Here's a stupid-simple example. of course, the compilers fail with (code snippet below image):

class Z {
    var common = 4
}

class A: Z {

}

class B: Z {

}

class C: Z {
    var specific: String? = "%"
}


let unknown = Z()

switch (unknown, unknown.common) {
case (let a as A, 4):
    break

case (let b as B, 4):
    break

case (let c as C, 4), let nonNilSpecific as? String:
    // use nonNilSpecific WITHOUT unwrap it within the case clousre
    break
default: break
}

Answer 1

When you use multiple patterns in a single case of a switch, they must bind all of the same variables.

Swift sees this line:

case (let c as C, 4), let nonNilSpecific as? String:

and thinks you're trying to match either (let c as C, 4) or let nonNilSpecific as? String. Those two choices bind different variables, so in the case body it is impossible to know which variables have been bound.

Perhaps you wanted something like this:

switch (unknown, unknown.common) {
case (let a as A, 4):
    break

case (let b as B, 4):
    break

case (let c as C, 4) where c.specific != nil:
    // force unwrap safe here, no closure needed
    let nonNilSpecific = c.specific!

default: break
}

---

Use an if:

let tuple = (unknown, unknown.common)
if case (let a as A, 4) = tuple {
    // use a
} else if case (let b as B, 4) = tuple {
    // use b
} else if case (let c as C, 4) = tuple, let nonNilSpecific = c.specific {
    // use c and nonNilSpecific
} else {
    // default case
}