Swift calling static methods: type(of: self) vs explicit class name

问题

In swift, an instance func can't call a static/class func without prefixing the method call with the class name. OR you can use type(of: self), e.g

class Foo {
    static func doIt() { }

    func callIt() {
        Foo.doIt() // This works
        type(of: self).doIt() // Or this

        doIt() // This doesn't compile (unresolved identifier)
    }
}

My question is, what's the difference here? Is it just a matter of coding style, or is there some difference e.g. static or dynamic dispatch going on?

If it is just coding style, what's the preferred style?

回答1:

There are two main differences.

1. The value of self inside the static method

The metatype that you call the static method on is available to you in the method as self (it's simply passed as an implicit parameter). Therefore if you call doIt() on type(of: self), self will be the dynamic metatype of the instance. If you call it on Foo, self will be Foo.self.

class Foo {
    static func doIt() {
        print("hey I'm of type \(self)")
    }

    func callDoItOnDynamicType() {
        type(of: self).doIt() // call on the dynamic metatype of the instance.
    }

    func classDoItOnFoo() {
        Foo.doIt() // call on the metatype Foo.self.
    }
}

class Bar : Foo {}

let f: Foo = Bar()

f.callDoItOnDynamicType() // hey I'm of type Bar
f.classDoItOnFoo()        // hey I'm of type Foo

This difference can be really important for factory methods, as it determines the type of instance you create.

class Foo {
    required init() {}

    static func create() -> Self {
        return self.init()
    }

    func createDynamic() -> Foo {
        return type(of: self).create()
    }

    func createFoo() -> Foo {
        return Foo.create()
    }
}

class Bar : Foo {}

let f: Foo = Bar()

print(f.createDynamic()) // Bar
print(f.createFoo())     // Foo

2. The dispatching of the static method

(Martin has already covered this, but I thought I would add it for the sake of completion.)

For class methods that are overridden in subclasses, the value of the metatype that you call the method on determines which implementation to call.

If called on a metatype that is known at compile time (e.g Foo.doIt()), Swift is able to statically dispatch the call. However, if you call the method on a metatype that isn't known until runtime (e.g type(of: self)), the method call will be dynamically dispatched to the correct implementation for the metatype value.

class Foo {
    class func doIt() {
        print("Foo's doIt")
    }

    func callDoItOnDynamicType() {
        type(of: self).doIt() // the call to doIt() will be dynamically dispatched.
    }

    func classDoItOnFoo() {
        Foo.doIt() // will be statically dispatched.
    }
}


class Bar : Foo {
    override class func doIt() {
        print("Bar's doIt")
    }
}

let f: Foo = Bar()

f.callDoItOnDynamicType() // Bar's doIt
f.classDoItOnFoo()        // Foo's doIt

回答2:

For a class method it makes a difference if the method is overridden in a subclass:

class Foo {
    class func doIt() {
        print("Foo doit")
    }

    func callViaClassname() {
        Foo.doIt()
    }

    func callViaTypeOf() {
        type(of: self).doIt()
    }
}

class Bar: Foo {
    override class func doIt() {
        print("Bar doit")
    }

}

Bar().callViaClassname() // Foo doit
Bar().callViaTypeOf() // Bar doit

This is also documented in "Types" in the Swift Language Reference:

You can use a type(of:) expression with an instance of a type to access that instance’s dynamic, runtime type as a value, ...

I don't know a difference for a static method (which is final and cannot be overridden in a subclass). Correction: See Hamish's answer for the difference in both static and class methods.

来源：https://stackoverflow.com/questions/42260337/swift-calling-static-methods-typeof-self-vs-explicit-class-name