is it possible to create a generic closure in Swift?

Probably you need something like this.

Type declaration:

typealias ResultClosure<T> = (ResultCode, String?, T?) -> Void

Function declaration:

func loginUser(userName: String, password: String, resultHandler: ResultClosure<TokenModel>?)

Usage:

    NetConnector.shared.loginUser(userName: userName ?? "", password: password ?? "") { (code, message, data) in
        self.display?.unlockScreen()
        if code == .success {
            if let activeToken = data {
                AppData.shared.userToken = activeToken
            }
            self.display?.showHome()
        } else {
            self.display?.showError(errorMessage: message)
        }
    }

No, because variables and expressions can't be generic. There are only generic functions and generic types.

---

To clarify: In some languages you can have types with a universal quantifier, like forall a. a -> a. But in Swift, types cannot have a universal quantifier. So expressions and values cannot be themselves generic. Function declarations and type declarations can be generic, but when you use such a generic function or an instance of such a generic type, some type (which could be a real type or a type variable) is chosen as the type argument, and thereafter the value you get is no longer itself generic.