Java parameter confusion

Question

In the code below why is it that o1.equals(o2); calls equals(Object o) not the equals(EqualsTest et) even though o1 and o2 are referencing objects of type EqualsTest!

public class EqualsTest {
      public static <T> boolean equalTest(T o1, T o2) {
          return o1.equals(o2);
      }
      public static void main(String[] args) {
          EqualsTest et1 = new EqualsTest();
          EqualsTest et2 = new EqualsTest();
          System.out.println(et1.equals(et2));
          System.out.println(equalTest(et1, et2));
      }
      public boolean equals(Object o) {
          if (o instanceof EqualsTest) {
              System.out.println("equals(Object)");
              return true;
          }
          return false;
      }
      public boolean equals(EqualsTest et) {
          System.out.println("equals(EqualsTest)");
          return this.equals((Object)et);
      }
}

You should never write an equals(X) method except for equals(Object), FYI. It'll only get you confused, and won't get used when you expect it to. — Louis Wasserman
– Louis Wasserman, Commented Nov 5, 2012 at 17:31
@LouisWasserman yes, this is where im confused on. I see methods written with Object params all the time, but if I knew I'm going to pass in an EqualsTest object, why can't I get specific and set the param to EqualsTest? — user133466
– user133466, Commented Nov 5, 2012 at 17:38
Because all the other code that uses equals -- like HashSet -- has to worry about the case where two objects of different types can be unequal, so it has to use the equals(Object) version, and the overload rules mean that it won't call your equals(EqualsTest) version. It's sort of weird at first, but it really is the only approach that makes sense in all cases. — Louis Wasserman
– Louis Wasserman, Commented Nov 5, 2012 at 18:38
@LouisWasserman does that mean if i declared public static <EqualsTest> boolean equalTest(EqualsTest o1, EqualsTest o2), then the program would call equals(EqualsTest et)? — user133466
– user133466, Commented Nov 5, 2012 at 18:41

Community · Accepted Answer · 2017-05-23 12:12:20Z

1

The compiler finds the corresponding method based on the declared type of the argument, not the most specific one. Since you didn't specify anything for thee T, it defaults to Object, as @nicholas.hauschild correctly points out.

edited May 23, 2017 at 12:12

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answered Nov 5, 2012 at 17:24

Volker Stolz

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8 Comments

nicholas.hauschild Over a year ago

What is the declared type one might ask? In this scenario, it is <T>, which we have no further information about, so the only thing we can infer is that it is an Object. +1

user133466 Over a year ago

@nicholas.hauschild but didn't we referenced et2 = new EqualsTest() so the compiler would know that it is equals(EqualsTest et) that should be called... that's where im confused

Volker Stolz Over a year ago

No, once you are in the static generic method, this knowledge is "lost" since the parameters are just typed as Object.

user133466 Over a year ago

@ShiDoiSi if what you said is true, then there's no point in writing code (ever?) with equals(SomeClass obj) every method that deals with a class param must be written in methodName(Object obj)

Volker Stolz Over a year ago

@user133466 It will call this method if the argument type matches exactly at compile time, e.g. because a variable is declared as that type, or an explicit cast.

|

Amit Deshpande · Accepted Answer · 2012-11-05 17:59:05Z

1

As I mentioned in the comment it is because of TypeErasure in java.

Check the byte code which is generated for equalTest. You can clearly see it will invoke method which has Object as parameter.

It is same as calling this.equals((Object)et) which will invoke Object method

// Method descriptor #15 (Ljava/lang/Object;Ljava/lang/Object;)Z
// Signature: <T:Ljava/lang/Object;>(TT;TT;)Z
// Stack: 2, Locals: 2
public static boolean equalTest(java.lang.Object o1, java.lang.Object o2);
0  aload_0 [o1]
1  aload_1 [o2]
2  invokevirtual java.lang.Object.equals(java.lang.Object) : boolean [18]

What you need is

public static <T extends EqualsTest> boolean equalTest(T o1, T o2) {
    return o1.equals(o2);
}

Now Lets check the generated byte code.

public static boolean equalTest(EqualsTest o1, EqualsTest o2);
0  aload_0 [o1]
1  aload_1 [o2]
2  invokevirtual EqualsTest.equals(EqualsTest) : boolean [18]
5  ireturn

As you can see compiler has changed Object to specific type that is EqualsTest because we have used Bounded Type so now if you invoke equalTest it will invoke method with equalTest as parameter

edited Nov 5, 2012 at 17:59

answered Nov 5, 2012 at 17:41

Amit Deshpande

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2 Comments

Volker Stolz Over a year ago

His confusion is that Java doesn't use the runtime-type of arguments for dispatching. But your answer is helpful in clarifying that the missing type instantiation for T turns into Object.

Volker Stolz Over a year ago

@user133466 You can look at the generated bytecode using javap: docs.oracle.com/javase/1.5.0/docs/tooldocs/windows/javap.html

Santosh Gokak · Accepted Answer · 2012-11-05 18:36:43Z

1

Since you have used Overloading the methods are linked at compile time and Java uses the less specific argument type for the polymorphic binding to the callee which is object in this case and not EqualsTest.

edited Nov 5, 2012 at 18:36

answered Nov 5, 2012 at 17:27

Santosh Gokak

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5 Comments

Volker Stolz Over a year ago

-1, least specific != most precise at compile time, which is what actually happens. (And I don't find that article particularly readable ;)

Santosh Gokak Over a year ago

What you mean by "most precise"?

Volker Stolz Over a year ago

To clarify: the compiler chooses Object, because this is the only type it has at compile type. If it has a "better" one like in @AmitD's answer, it will use that one, not a least specific as you are conjecturing (which would still be Object). Or did you mean "...the less specific..."?

Santosh Gokak Over a year ago

i meant the less specific the compiler know of. Is that correct?

Volker Stolz Over a year ago

Try to elaborate your answer...this is about precision (literally), so also some care in formulating is required (as you rightfully pointed out before).

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