Why does '<', '> ' work with wrapper classes while '==' doesn't ? [duplicate]

Solution 1:

The <, >, <= and >= operators are only defined for primitive types. Therefore using them on wrapper types causes the compiler to unbox the objects into primitives.

This means

System.out.println(a>=b);

is equivalent to

System.out.println(a.intValue()>=b.intValue());

However, the == and != operators exist for both primitive types and reference types, so using them to compare two objects of primitive wrapper types compares the references instead of the primitive values they wrap.

As Holger commented, comparison of object references with == and != existed in the Java language before auto-boxing and auto-unboxing were introduced, but comparison with <, >, <= and >= was not supported for any reference types before auto-unboxing was introduced.

This means that in the early days of Java, the a>=b of your code snippet would not pass compilation (since a and b are not primitive numeric types). On the other hand your a==b snippet would still pass compilation and return false.

Changing the behavior of == and != for reference types that happen to be wrappers of numeric primitives would change the behavior of existing code, thus breaking backwards compatibility, which is probably the reason it wasn't done.