Some confusing cases of method overloading – Part 3

Following this series of posts related to method overloading in Java, now we will see how will the compiler behave when calling an overloaded method which actual parameter is a result of the conditional ? operator.

First of all, in the case you missed the previous posts, here you have the links to the first part, and the second part.

Let’s consider the following example program:

import java.util.Random;

public class OverloadPrimitiveTest {

  public void method(int a) {
    System.out.println("method(int)");
  }

  public void method(Integer b) {
    System.out.println("method(Integer)");
  }

  public static void main(String[] args) {
    OverloadPrimitiveTest o = new OverloadPrimitiveTest();

    o.method(1); // call method(int)
    o.method(new Integer(1)); // call method(Integer)
    o.method((short) 1); // call method(int)
    o.method(null); // call method(Integer)

    Random random = new Random();
    o.method((random.nextBoolean()) ? 1 : new Integer(2)); // call method(int), ambiguous
  }

}

The first two calls are straightforward to digest, they call the intuitively correct overloaded method since their actual parameters in the method call expression are those specific ones.

The third call seems strange at first sight but it is OK, since Java can safely widen a primitive short to an int (no overflow can happen), so this resolve to the overloaded method which takes a primitive int parameter. The fourth call is very intuitive also, because a null value can be assigned to reference types, not a primitive type, so using a null as the actual parameter resolves intuitively to the method that takes an Integer object.

The fifth call is what I consider ambiguous.

Why is it ambiguous?

The last call does a couple of things with its actual parameter. First of all, the actual parameter is taken after the evaluation of the conditional ? operator, and the type returned by the conditional operator is the primitive int, because the specification says that it will unbox its second and third operand as necessary (JLS 15.25.)

However, comparing the fifth invocation of the overloaded method with the first and second ones, it is confusing at first sight which one of the methods will be selected at compile-time.

An example using references

The same confusing case may happen with references to objects too, since it is not necessary the presence of a boxing or unboxing operation for the anomalous case to appear. Just like the primitive int type can be widened to long or float, a reference type can be widened to some superclass in the class herarchy, or some implemented interface. Here is an example:

import java.applet.Applet;
import java.awt.Panel;
import java.util.Random;

import javax.swing.JApplet;

public class OverloadPrimitiveTest2 {

 public void method(Panel p) {
   System.out.println("Panel");
 }

 public void method(JApplet a) {
   System.out.println("JApplet");
 }

 public static void main(String[] args) {
   OverloadPrimitiveTest2 o = new OverloadPrimitiveTest2();

   o.method(new Panel()); // prints Panel
   o.method(new JApplet()); // prints JApplet

   o.method((new Random().nextBoolean()) ? new Applet() : new JApplet()); // prints Panel
 }

}

Here, JApplet is a subclass of Applet, which in turn is a subclass of Panel. In the third method invocation the conditional ? operator decides that the type of the expression is Applet, since JApplet is widened to Applet without any problem. In turn, the best fit for the method invocation is calling method(Panel) since an expression of type Applet can be widened to type Panel. Again it is confusing the intention of the programmer since both methods may have been selected (if not for the conditional ? operator.)

A note on code checking tools

The usual code analysis tools, like FindBugs, PMD, and (now free) Google’s CodePro AnalytiX are happy to detect and mark the use of the conditional ? operator. I find that informative but more often than not, it marks a false positive, that is to say that not necessarily I believe that using the conditional ? operator is wrong: it has its use to make the program somewhat easier to read. However, I do think that the conditional ? operator is too smart in trying to reduce the two evaluated expressions into a single type expression: sometimes it widens too much and useful type information is lost in the oblivion.

As a side note, CodePro AnalytiX has a rule indicating that there is an overloaded method and recommends that either rename the method (to avoid the overloading) or change the number of parameters. I actually didn’t find this recommendation very useful, as the intention of an overloaded method is actually that by design it should have the same name (because they perform the same process for a different type) and it would by unnatural to add an extra parameter just for disambiguate the method invocation at compile-time.

Conclusions

The conditional ? operator is tricky, because it is too smart and infer things that most of the time we will not be aware of. In addition, combine method overloading with this operator and we will certainly have some fun time hunting bugs. However, I don’t personally think nor that the conditional ? operator should be avoided (as suggested by code analysis tools) nor method overloading, though I do believe you must not mix them.

You can distill a rule from this. Note that in the conditional ? operator, the types of the second and third operands are different and when considered in isolation, each one of them would make the compiler choose a different overloaded method to invoke. So a simple rule of thumb to remember would be:

Don’t use the conditional ? operator as an actual parameter of an overloaded method if the second and third operands have different types.

I wish some day I could see a rule implemented on those code analysis tools to detect this tricky cases.

Some confusing cases of method overloading – Part 2

Some days ago I posted about an example program which called an overloaded method where its parameters are final variables. This time I’ll call an overloaded constructor where its actual parameter is a null literal. As per the Java Language specification, overloaded constructor resolution is identical to overloaded method resolution, so for methods these same concepts apply.

Consider this example, extracted from the (very good and entertaining) book Java Puzzlers, which is a perfectly legal Java program:

public class Confusing {

  private Confusing(Object o) {
    System.out.println("Object");
  }

  private Confusing(double[] dArray) {
    System.out.println("double array");
  }

  public static void main(String[] args) {
    new Confusing(null); // prints double array
  }

}

Here, the class Confusing provides an overloaded constructor: one of them take an Object as a parameter and the other an array. Now, as the Java Language Specification says in its Chapter 10: arrays are objects, are dynamically created, and can be assigned to variables of type Object. All methods of the class Object can be invoked on an array.

So we have that double[] is a subclass of Object, and we try to create an instance of the class Confusing with its actual parameter being a null literal. Because null can be assigned to a double[] array, and to an Object instance, then both methods are applicable, however as per the Java Language Specification, the overloaded method to be called is the most specific one, and because an array is a subclass of Object, then the most specific is the Confusing(double[]) constructor, so the compiler chooses this constructor and then the program compiles without error on Java 6 update 20. Tools like FindBugs and PMD didn’t find this weird, as no warning was emitted from them.

Is it ambiguous or not?

Clearly the above program is perfectly legal…for the compiler. For me, and I believe many others average Joe programmers it is confusing. Why it is legal when at first glance it seems it is ambiguous? My rule of thumb is that code must be easy to read, because code is read more times than written. Also, for productivity purposes the compiler doesn’t count as somebody that read code, so why write perfectly well-formed code that is difficult to grasp at first sight?

The code is ambiguous because a null literal could be used as actual parameter for both constructors, it is not clear which one it will choose to call. Whether or not the Java compiler actually has very deterministic and definite rules to disambiguate the call, it is irrelevant to the human (and error prone) reader, and therefore it is sound to emit a Warning message here.

How to disambiguate

Casting the null parameter to the actual type you intent to refer to is the simplest way to disambiguate the call. No question which constructor you are referring to now:

new Confusing((Object)null);
new Confusing((String[])null);

or you can define a variable with the right type:

Object x = null;
new Confusing((x);
double[] y = null;
new Confusion(y);

Conclusions

The rules of overloaded method resolution are complicated to understand, and “smart” code that depends on those rules makes the program harder to read. I would personally like to see FindBugs or PMD add some rule to detect this kind of situation. As said in the first part of this series of posts, why don’t we use the Java type system to the fullest?

Stay tuned for more interesting overloaded examples!

Some confusing cases of method overloading

Some time ago I was starting to wonder how the Java compiler actually worked, because some programs I was reading at that time compiled without errors but looked like something was out-of-place, particularly some example programs which included overloaded methods.

Method overloading is basically a feature offered by a programming language allowing to create several methods with the same name, but different parameters type or output type. Java is one of such languages.

Java defines at compile time which method will be actually called when encountered with a method invocation. The rules that are checked by the compiler to take such a decision are rather complex and can be found in the Java Language Specification for the intrigued reader. Those rules may or may not be perfect but are the ones that apply to every Java program, so proposing a change on them produces the very undesirable effect of making every Java program to potentially fail to compile. However, the compiler may be modified to emit a Warning when something looks suspicious, even if the program compiles correctly. In this post I’ll show you some of those cases:

Calling an overloaded method where the actual parameter is a final variable

class A {
}

class B extends A {
}

public class OverloadingTest {

  public void method(A a) {
    System.out.println("method(A)");
  }

  public void method(B b) {
    System.out.println("method(B)");
  }

  public static void main(String[] args) {
    OverloadingTest o = new OverloadingTest();

    A p = new B();
    o.method(p); // call method(A)

    o.method(new B()); // call method(B)

    final A p2 = new B();
    o.method(p2); // call method(A)
  }
}

In this example, the first and second calls are intuitively resolved to what the programmer expect, however the third call is open to some controversy. Since the actual method called by the program is resolved at compile time, the Java compiler checks the actual parameters passed to the method call against its static type, not its dynamic type.

This makes sense since at compile time, the compiler can not determine actually which will be the dynamic type of a variable when the program is running. In the third call, the p2’s static type is A and its dynamic type is B. However the p2 variable is marked as final, meaning it can not change its type during its lifetime. In other words, p2 is instantiated as having the B type, and its dynamic type can not change…ever.

IMO, this seems to me like a programmer error on the intent of the variable p2. It is a B, and it will never by any other type? or it is an A? Because based on that decision the compiler will call one or the other overloaded method. I believe the compiler can check on this and show a Warning. I checked this code compiling against Java 6 Update 20 and no warning, also FindBugs and PMD didn’t detect this as a code smell either.

There are some ways to disambiguate the conflicting line of code:

  • Cast the variable to the more specific type:
    final A p2 = new B();
    o.method((B)p2);
  • Remove the final modifier (meaning that its dynamic type could change):
    A p2 = new B();
    o.method((B)p2);
  • Keep the final modifier, but changing the static type to one that indeed lead the compiler to resolve to one (and only one) matching method:
    final B p2 = new B();
    o.method(p2);
    // or
    final A p2 = new A();
    o.method(p2);

Dealing with blank final declarations

Consider the following code excerpt:

final A a;
if (someCondition) {
  a = new A();
}
else {
  a = new B();
}
o.method(a);

There is no obvious way of detecting the issue here (at least not without a flow analysis) so a conservative thing to do is to raise the warning anyway, so the programmer may take out either the final keyword, or refactor the code like this:

if (someCondition) {
  final A a = new A();
  o.method(a);
}
else {
  final B b = new B();
  o.method(b);
}

This time, PMD somewhat points to a problem here, since it warns about the use of final (blank) local variables (AvoidFinalLocalVariable Rule.)

The conditional ? operator

Seems counter intuitive that if

o.method(new B());

call method(B), the following two actually resolve to method(A):

o.method((false) ? new A() : new B());
o.method((false) ? (A)null : new B());

and the following will call method(B):

o.method((false) ? null : new B());

This is so because of the resulting expression type produced by the ? operator.

As both the second and third operand of the ? operator are by all means constant expressions, IMO they should be treated just with the same semantics as if a blank final variable had been used instead:

final A a;
if (condition)
  a = expressionReturningA;
else
  a = expressionReturningB;
o.method(a);

Then, the same reasoning as before applies.

IMO, using the ? operator to resolve an expression for use as an actual parameter in an overloaded method call should be avoided if possible, it’s a code smell, and as such, tools must warn the programmer. Since the ? operator has only one type (which depends on the second and third operand), the only case when it may not emit a warning is if both operands reduce to a type that uniquely resolve to only one (and the same) method.

Conclusions

The interaction between overloaded methods and different kind of expressions in Java is tricky. On the one hand, method resolution rules are very complicated, but reasoning about expressions that represent constants or non changing values (such as final variables) should allow the programmer to think about the code in abstract terms, just as any mathematical formula would behave. Why bother to program in a strongly typed language if we can not reason about the program in an intuitive fashion? Why not make the tools to emit a warning when we write somewhat ambiguous code? Why can’t we use the Java type system to the fullest?

In the upcoming weeks I’ll post some more obscure examples.

Calling an overloaded method where the actual parameter is a final variable

Hello world!

Hello!! This is my first post. I’ll be writing about things that usually are related to the software development area.  I must state clear that this blog will have many personal opinions, and frecuently some source code to read.

By now I leave you with this:

public class Test {
  public static final void main(String[] argv) {
    System.out.println("Hello world!");
  }
}