Swift Switch 访问 'case' 内的嵌套属性

Question

遵循此代码段。我试图了解是否有可能在 Switch 语句中访问对象的嵌套属性，而无需在“案例”本身内展开属性(避免不必要的闭包)。 这是一个愚蠢的简单例子。当然，编译器会失败(图片下方的代码片段):

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

当您在 switch 的单个案例中使用多个模式时，它们必须绑定(bind)所有相同的变量。

Swift 看到这一行:

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

并认为您正在尝试匹配 (let c as C, 4) 或 let nonNilSpecific as？字符串。这两个选择绑定(bind)了不同的变量，因此在案例主体中不可能知道绑定(bind)了哪些变量。

也许你想要这样的东西:

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
}

---

使用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
}