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Unit 15: Method Invocation

Learning Objectives

After this unit, the student should:

  • understand the two-step process that Java uses to determine which method implementation will be executed when a method is invoked
  • understand that Class Methods do not support dynamic binding

How does Dynamic Binding work?

We have seen that, with the power of dynamic binding and polymorphism, we can write succinct, future-proof code. Recall that example below, where the magic happens in Line 4. The method invocation curr.equals(obj) will call the corresponding implementation of the equals method depending on the run-time type of curr.

contains v0.1 with Polymorphism
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boolean contains(Object[] array, Object obj) {
  for (Object curr : array) {
    if (curr.equals(obj)) {
      return true;
    }
  }
  return false;
}

How does dynamic binding work? To be more precise, when the method equals is invoked on the target curr, how does Java decide which method implementation is this invocation bound to? While we have alluded to the fact that the run-time type of the target curr plays a role, this is not the entire story. Recall that we may have multiple versions of equals due to overloading. So, Java also needs to decide, among the overloaded equals, which version of equals this particular invocation is bound to.

This unit elaborates on Java's decision process to resolve which method implemented in which class should be executed when a method is invoked. This process is a two-step process. The first occurs during compilation; the second during run time.

During Compile Time

During compilation, Java determines the method descriptor of the method invoked, using the compile-time type of the target.

For example, in the line

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curr.equals(obj)
above, the target curr has the compile-time type Object.

Let's generalize the compile-time type of the target to \(C\). To determine the method descriptor, the compiler searches for all methods that can be correctly invoked on the given argument.

In the example above, we look at the class Object, and there is only one method called equals. The method can be correctly invoked with one argument of type Object.

What if more than one methods can correctly accept the argument? In this case, we choose the most specific one. Intuitively, a method \(M\) is more specific than method \(N\) if the arguments to \(M\) can be passed to \(N\) without compilation error. For example, let's say a class Circle implements:

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boolean equals(Circle c) { .. }

@Override
boolean equals(Object c) { .. }

Then, equals(Circle) is more specific than equals(Object). Every Circle is an Object, but not every Object is a Circle. Let's try to understand this using our definition of "more specific" above.

Consider the second part of the definition, "if the arguments to \(M\) can be passed to \(N\) without compilation error". We need to find which arguments can be accepted by the methods we wish to compare. In the case of equals(Circle), it can accept an argument of compile-time type Circle (and all its subclasses), but not an argument of compile-time type Object. On the other hand, equals(Object) can accept an argument of compile-time type Object and all its subclasses, including Circle.

Therefore, since all arguments to equals(Circle) can be passed to equals(Object) without compilation error, we say equals(Circle) is more specific than equals(Object).

There is also the possibility that when comparing two methods, none of the two methods is more specific than the other. For instance, given S1 <: T and S2 <: T, foo(S1) is not more specific than foo(S2) and foo(S2) is not more specific than foo(S1). If the Java compiler fails to determine a single most specific method, it will throw a compilation error.

Otherwise, once the Java compiler determines the most specific method, it stores the method's descriptor (return type and signature) in the generated bytecode.

In the example above, the method descriptor boolean equals(Object) will be stored in the generated binaries. Note that it does not include information about the class that implements this method. The class from which this method implementation will be taken is determined in Step 2 during run time.

During Run Time

During execution, when a method is invoked, the method descriptor from Step 1 is first retrieved. Then, the run-time type of the target is determined. Let the run-time type of the target be \(R\). Java then looks for an accessible method with the matching descriptor in \(R\). If no such method is found, the search will continue up the class hierarchy, first to the parent class of \(R\), then to the grand-parent class of \(R\), and so on, until we reach the root Object. The first method implementation with a matching method descriptor found will be the one executed.

For example, let's consider the invocation in the highlighted line below again:

v0.1 without Polymorphism
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boolean contains(Object[] array, Object obj) {
  for (Object curr : array) {
    if (curr.equals(obj)) {
      return true;
    }
  }
  return false;
}

Let's say that curr points to a Circle object during run-time. Suppose that the Circle class does not override the method equals in Object. As a result, Java cannot find a matching method descriptor boolean equals(Object) in the method Circle. It then looks for the method in the parent of Circle, which is the class Object. It finds the method Object::equals(Object) with a matching descriptor. Thus, the method Object::equals(Object) is executed.

Now, suppose that Circle overrides the method Object::equals(Object) with its own Circle::equals(Object) method. Since Java starts searching from the class Circle, it finds the method Circle::equals(Object) that matches the descriptor. In this case, curr.target(obj) will invoke the method Circle::equals(Object) instead.

Invocation of Class Methods

The description above applies to instance methods. Class methods, on the other hand, do not support dynamic binding. The method to invoke is resolved statically during compile time. The same process in Step 1 is taken, but the corresponding method implementation in class \(C\) will always be executed during run-time, without considering the run-time type of the target.