Unit 15: Method Invocation

Learning Objectives

Students 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 code¹. 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
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-part 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

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 first searches for all methods that are accessible and have the same name.

In this case, if we look at the class Object, there is only one method called equals. Additionally, Object does not have a superclass, so boolean equals(Object) is the only accessible methods with the name equals.

If there is none, then the compiler throws an error. Otherwise, the compiler checks for compatible methods. By compatible, we mean all methods that can be correctly invoked on the given argument.

In the example above, the method can be correctly invoked with one argument of type Object.

What if there are multiple methods that can correctly accept the argument? In this case, we choose the most specific one.

More Specific

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:

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 what 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.

Now we simply have to test if equals(Circle) can accept whatever can be accepted by equals(Object) and vice versa. So we test equals(Circle) and attempt to pass Object and Circle as argument (since equals(Object) can accept both Object and Circle). It will pass on Circle but it will fail on Object.

Similarly, we test equals(Object) and attempt to pass Circle (since equals(Circle) can only accept Circle). There is definitely no compilation error here. So now we have seen that if we set \(M\) as the method equals(Circle) and \(N\) as equals(Object),

the arguments to \(M\) (i.e., equals(Circle), the argument is Circle) can be passed to \(N\) (i.e., equals(Object) can accept Circle) without compilation error.

Therefore, equals(Circle) is more specific than equals(Object).

There is a possibility that comparing only 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 to take this method implementation from will be 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.

Matching Method

A method \(M\) matches the method \(N\) if they have the same method signature and the returns type of \(M\) is a subtype of the method \(N\). Recap that \(T\) <: \(T\).

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

contains v0.1
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 can't 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.

Steps

The information above may be difficult to parse and understand. So let's try to distill its essence into actionable steps. In this example, we want to figure out the method invoked on obj.foo(arg).

Compile-Time Step

Determine the compile-time type of obj (i.e., CTT(obj)).
Determine the compile-time type of arg (i.e., CTT(arg)).
Determine all the methods with the name foo that are accessible in CTT(obj).
- This includes the parent of CTT(obj), grand-parent of CTT(obj), and so on.
- The access modifiers are appropriate.
Determine all the methods from Step 3 that are compatible with CTT(arg).
- Correct number of parameters.
- Correct parameter types (i.e., supertype of CTT(arg)).
Determine the most specific method from Step 4.
- If there is no most specific method, fail with compilation error.
- Otherwise, record the method descriptor.

Run-Time Step

Retrieve the method descriptor obtained from compile-time step.
Determine the run-time type of obj (i.e., RTT(obj)).
Starting from RTT(obj), find the first method that match the method descriptor as retrieved from Step 1.
- If not found, check in the parent of RTT(obj).
- If not found, check in the grand-parent of RTT(obj).
- :
- If not found, check in the root Object.
- If not found, run-time error.

To see the steps in action, please follow the examples below.

Example

Although the steps above are actionable, it is still instructive to at least see how the steps are carried out. We will be using the following classes in our example.

class U {
  void foo(T t) { }
  void foo(U u1, U u2) { }
}

class T extends U {
  void foo(S s) { }
}

class S extends T {
  void foo(U u) { }
}

Consider the following variables.

U u = new T();
S s = new S();

In the last example, we will also show a tabular method that simplifies the steps.

u.foo(s)u.foo(u)s.foo(s)s.foo(s) Tabular

Compile-Time Step

CTT(obj) = U
CTT(arg) = S
foo includes foo(T), foo(U, U)
S can be accepted by foo(T)
The most specific is void foo(T)

Run-Time Step

Descriptor = void foo(T)
RTT(obj) = T
Check method from T
- void T::foo(T)
- void U::foo(T)

Compile-Time Step

CTT(obj) = U
CTT(arg) = U
foo includes foo(T), foo(U, U)
U cannot be accepted by any method from Step 3.

Compilation-Error

Compile-Time Step

CTT(obj) = S
CTT(arg) = S
foo includes foo(U), foo(S), foo(T), foo(U, U)
S can be accepted by foo(U), foo(S), foo(T)
The most specific is void foo(S)

Run-Time Step

Descriptor = void foo(S)
RTT(obj) = S
Check method from S
- void S::foo(S)
- void T::foo(S)

Compile-Time Step

`foo`	Accept `foo(S)`	Most specific
`foo(U)`
`foo(S)`
`foo(T)`
`foo(U, U)`		- (can be ignored)

Run-Time Step

Class	Has `void foo(S)`?
`S`
`T`
`U`	- (can be ignored)

Hopefully the 3 examples above are informative. We cannot cover all possibilities but the steps should provide guidance on what to do when new situation arise. In recitation we will connect this with class diagram so that we can understand the steps visually.

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.

Bad Practice

To show that invocation of class methods is not via dynamic binding, we have to use example that are basically frowned upon. Do not follow this bad practice in your coding.

class T {
  static int f() {
    return 1;
  }
}

class S extends T {
  static int f() {
    return 2;
  }
}

Typically, we will invoke the static method f using either T.f() or S.f(). Unfortunately, to show that invocation of class method is using static binding, we have to use the instance rather than the class. Which leads us to the unorthodox code snippet below.

T t = new S(); // compile-time type is T, run-time type is S
System.out.println(t.f()); // what is the expected result?

Unless the future requires us to add functionality to every single classes. ↩