How to choose overloaded method by argument type?

Paweł Sosnowski :

I'm trying to overload my method depending on the argument type:

FieldError extends ObjectError.

private String getMessage(final FieldError fieldError) {
    return String.format("%s %s", fieldError.getField(), fieldError.getDefaultMessage());
}

private String getMessage(final ObjectError objectError) {
    return objectError.getDefaultMessage();
}

Now, if I have a List<ObjectError> and call above method for elements of that List, regardless of which type (ObjectError or FieldError) the elements on the List are,
getMessage(final ObjectError objectError) will always be called.

List<ObjectError> errors;
getMessage(errors.get(0));

Why is it behaving this way? If it's just how it works, is there any chance not using if's and instanceof's?

Eran :

Method overloading resolution is performed at compile time, when only the compile-time type is known. Therefore, the compile-time type of the parameters passed to the method call determine which of the overloaded methods is chosen.

Since you are passing elements of a List<ObjectError> to the method call, String getMessage(final ObjectError objectError) is always chosen, and the runtime type of the instances stored in that List doesn't make a difference.

One way to solve your problem is to move the getMessage() method to ObjectError class, and override it in the FieldError sub-class.

Instead of

getMessage(errors.get(0))

you'll call

errors.get(0).getMessage()

This will work, since method overriding is resolved at runtime.