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What is the difference between * and? I read in another post that * is used to denote that the package is deprecated and that you should use? instead. But I’ve also read that if we want to replace a method in a package with one in a class we should use? and not *, even though it would change the signature of the method (it would actually be the signature of the class, as it gets replaced by it’s method, not the package one). Is it just a convention and if so, what is it? If it’s not, what is the reason for this?

A:

Your definition of * is correct, but what you read about the difference is misleading, and it is not necessary to use the * if you don’t want to.

Using the * is just a way to explicitly tell the type checker that a type is deprecated, not because the old methods were incompatible with the new methods. It is the easiest way, because if the * does not say what the type is, then the type checker may not realize that the old type is deprecated. This means that if you have to change the type anyway, you’ll have to explicitly use the * to make sure that you don’t break anything.

If you don’t want to use the *, then you can use the new type name directly.

What you are probably reading is that there is a problem if you have a method or type that has a signature that is deprecated. This is not true, because in this case the type checker will not be able to recognize the deprecated type name, and it will fail to compile. For instance, if you have

object typeOfInt;

void myFunction(int) {

typeOfInt.method();

}

then the typeOfInt.method() call will fail to compile. The same thing is also true for classes, so using the new type name is a valid solution for that.

But if you have a method that has the same name, but not the same signature, then there is no problem. For instance, if you have

myFunction(42);

the two calls will be equivalent, because they have the same name, so there is no problem.

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