Are inout variables of protocol type prohibited?

Question

The following code:

protocol SomeProtocol {}
class SomeClass: SomeProtocol {}

private func doSomethingWith(inout someVar: SomeProtocol) {}

private var someGlobalVar = SomeClass() // inferring SomeClass's type

doSomethingWith(&someGlobalVar)

produces the following error:

Cannot invoke 'doSomethingWith' with an argument list of type '(inout SomeClass)'

Changing the penultimate line to private var someGlobalVar: SomeProtocol = SomeClass() resolves the error.

Subj.

Answer 1

When you assign a SomeClass instance to a variable while declaring, the variable type is inferred to be SomeClass. The same as writing

private var someGlobalVar: SomeClass = SomeClass()

However, when passing to an inout parameter, the function can assign another instance to that variable, e.g.

private func doSomethingWith(inout someVar: SomeProtocol) {
    someVar = OtherClass()
}

Now you have a type mismatch. The error you are seeing is Swift preventing you getting a similar problem.

In other words: if you are passing a variable to a function and you know that the function can assign any instance adopting SomeProtocol to that variable, you have to use a variable that can actually hold any instance adopting SomeProtocol:

private var someGlobalVar: SomeProtocol