Compilation error with == operator

Question

I've isolated the error to this line: string.getClass() == jojo.getClass() Shouldn't this line create two Class objects and then check if they (as in the two references) point to the same object? Rather than returning a value of false, the code won't run.

public class Tester 
{
    public static void main(String[] args) 
    {
        OreoJar jojo = new OreoJar(0);
        OreoJar momo = new OreoJar(1);
        String string = "Hello";

        if (momo.getClass() == jojo.getClass())
        {
            System.out.println("Momo and jojo are of the same class");
        }

        if (string.getClass() == jojo.getClass())
        {
            System.out.println("String and jojo are of the same class");
        }
    }
}

public class OreoJar 
{
    int oreos;

    public OreoJar(int oreos)
    {
        this.oreos = oreos;
    }

    public void count()
    {
        System.out.println(oreos + " oreos in this jar!");
    }
}

This comment is kind of hidden and I think its worth mentioning since it makes the most sense to a beginner (such as myself)

-According to the JLS "It is a compile-time error if it is impossible to convert the type of either operand to the type of the other by a casting conversion" so two references of types A and B can be compared if, and only if, either A can be cast to B or B can be cast to A. – Patricia Shanahan

Answer 1

I agree OP should quote the compilation error.

Anyway the compilation error is quite obvious when anyone actually does a compilation.

The error is:

Tester.java:15: incomparable types: java.lang.Class<capture#125 of ? extends java.lang.String> and java.lang.Class<capture#29 of ? extends OreoJar>
    if (string.getClass() == jojo.getClass()){
                          ^

Reason seems obvious.

From Javadoc of Object.getClass():

The java.lang.Class object that represents the runtime class of the
object. The result is of type Class<? extends X> where X is the
erasure of the static type of the expression on which getClass is
called.

That means, an String instance is going to return a reference to Class<? extends String>, while an OreoJar instance is going to return reference to Class<? extends OreoJar>

The two types are simply not compatible, as the compiler knows that there is no chance that any type that extends String can be a type extends OreoJar. So comparison is going to cause compilation error.

---

A bit off topic but I think worth mentioning, you said:

Shouldn't this line create two Class objects and then check if they point to the same object

I think it is better to have clearer understanding. It is not going to "create" two Class objects. getClass() is going to return you a reference to Class object. And, it is always a reference that can point to an object, not object that point to object (it sounds weird too)

Answer 2

I think the reason it won't compile is due to the fact that Class has generic component. Try using momo.getClass().equals(jojo.getClass()) And you might also try comparing the canonical names of the classes for a similar effect: momo.getClass().getCanonicalName().equals(jojo.getClass().getCanonicalName())

Answer 3

getClass() returns an instance of a Class. getClass().getName() returns a string. The String.equals(otherString) method is the correct way to compare Strings for equality.