Duck Typing vs. Type Erasure

Consider the following C++ class.

#include <iostream>

template <typename T>
struct Caller {
  const T callee_;
  Caller(const T callee) : callee_(callee) {}
  void go() { callee_.call(); }
};

Caller can be parameterized to any type so long as it has a call() method. For example, introduce two types, Foo and Bar.

struct Foo {
  void call() const { std::cout << "Foo"; }
};

struct Bar {
  void call() const { std::cout << "Bar"; }
};

int main() {
  Caller<Foo> foo{Foo()};
  Caller<Bar> bar{Bar()};
  foo.go();
  bar.go();
  std::cout << std::endl;
  return 0;
}

This code compiles cleanly and, when run, emits “FooBar”. This is an example of duck typing — i.e., “If it looks like a duck, swims like a duck, and quacks like a duck, then it probably is a duck.” Foo and Bar are unrelated types. They have no common inheritance, but by providing the expected interface, they both work with with Caller. This is a special case of polymorphism.

Duck typing is normally only found in dynamically typed languages. Thanks to templates, a statically, strongly typed language like C++ can have duck typing without sacrificing any type safety.

Java Duck Typing

Let’s try the same thing in Java using generics.

class Caller<T> {
    final T callee;
    Caller(T callee) {
        this.callee = callee;
    }
    public void go() {
        callee.call();  // compiler error: cannot find symbol call
    }
}

class Foo {
    public void call() { System.out.print("Foo"); }
}

class Bar {
    public void call() { System.out.print("Bar"); }
}

public class Main {
    public static void main(String args[]) {
        Caller<Foo> f = new Caller<>(new Foo());
        Caller<Bar> b = new Caller<>(new Bar());
        f.go();
        b.go();
        System.out.println();
    }
}

The program is practically identical, but this will fail with a compile-time error. This is the result of type erasure. Unlike C++’s templates, there will only ever be one compiled version of Caller, and T will become Object. Since Object has no call() method, compilation fails. The generic type is only for enabling additional compiler checks later on.

C++ templates behave like a macros, expanded by the compiler once for each different type of applied parameter. The call symbol is looked up later, after the type has been fully realized, not when the template is defined.

To fix this, Foo and Bar need a common ancestry. Let’s make this Callee.

interface Callee {
    void call();
}

Caller needs to be redefined such that T is a subclass of Callee.

class Caller<T extends Callee> {
    // ...
}

This now compiles cleanly because call() will be found in Callee. Finally, implement Callee.

class Foo implements Callee {
    // ...
}

class Bar implements Callee {
    // ...
}

This is no longer duck typing, just plain old polymorphism. Type erasure prohibits duck typing in Java (outside of dirty reflection hacks).

Signals and Slots and Events! Oh My!

Duck typing is useful for implementing the observer pattern without as much boilerplate. A class can participate in the observer pattern without inheriting from some specialized class or interface. For example, see the various signal and slots systems for C++. In constrast, Java has an EventListener type for everything:

A class concerned with many different kinds of events, such as an event logger, would need to inherit a large number of interfaces.

Have a comment on this article? Start a discussion in my public inbox by sending an email to ~skeeto/public-inbox@lists.sr.ht [mailing list etiquette] , or see existing discussions.

This post has archived comments.

null program

Chris Wellons

wellons@nullprogram.com (PGP)
~skeeto/public-inbox@lists.sr.ht (view)