Is the 'override' keyword just a check for a overridden virtual method?

Question

As far as I understand, the introduction of override keyword in C++11 is nothing more than a check to make sure that the function being implemented is the overrideing of a virtual function in the base class.

Is that it?

Solution 1

That's indeed the idea. The point is that you are explicit about what you mean, so that an otherwise silent error can be diagnosed:

struct Base
{
    virtual int foo() const;
};

struct Derived : Base
{
    virtual int foo()   // whoops!
    {
       // ...
    }
};

The above code compiles, but is not what you may have meant (note the missing const). If you said instead, virtual int foo() override, then you would get a compiler error that your function is not in fact overriding anything.

Solution 2

Wikipedia quote:

The override special identifier means that the compiler will check the base class(es) to see if there is a virtual function with this exact signature. And if there is not, the compiler will error out.

http://en.wikipedia.org/wiki/C%2B%2B11#Explicit_overrides_and_final

Edit (attempting to improve a bit the answer):

Declaring a method as "override" means that that method is intended to rewrite a (virtual) method on the base class. The overriding method must have same signature (at least for the input parameters) as the method it intends to rewrite.

Why is this necessary? Well, the following two common error cases are prevented:

1. one mistypes a type in the new method. The compiler, unaware that it is intending to write a previous method, simply adds it to the class as a new method. The problem is that the old method is still there, the new one is added just as an overload. In this case, all calls towards the old method will function just as before, without any change in behavior (which would have been the very purpose of the rewriting).

2. one forgets to declare the method in the superclass as "virtual", but still attempts to re-write it in a subclass. While this will be apparently accepted, the behavior won't be exactly as intended: the method is not virtual, so access through pointers towards the superclass will end calling the old (superclass') method instead of the new (subclass') method.

Adding "override" clearly disambiguates this: through this, one is telling the compiler that three things are expecting:

1. there is a method with the same name in the superclass
2. this method in the superclass is declared as "virtual" (that means, intended to be rewritten)
3. the method in the superclass has the same (input*) signature as the method in the subclass (the rewriting method)

If any of these is false, then an error is signaled.

^{* note: the output parameter is sometimes of different, but related type. Read about covariant and contravariant transformations if interested.}

Solution 3

Found "override" is useful when somebody updated base class virtual method signature such as adding an optional parameter but forgot to update derived class method signature. In that case the methods between the base and the derived class are no longer polymorphic relation. Without the override declaration, it is hard to find out this kind of bug.

Solution 4

Yes, this is so. It's a check to make sure one doesn't try an override and mess it up through a botched signature. Here's a Wiki page that explains this in detail and has a short illustrative example:

http://en.wikipedia.org/wiki/C%2B%2B11#Explicit_overrides_and_final

Solution 5

C++17 standard draft

After going over all the override hits on the C++17 N4659 standard draft, the only reference I can find to the override identifier is:

5 If a virtual function is marked with the virt-specifier override and does not override a member function of a base class, the program is ill-formed. [ Example:

struct B {
  virtual void f(int);
};

struct D : B {
  virtual void f(long) override; // error: wrong signature overriding B::f
  virtual void f(int) override;  // OK
}

— end example ]

so I think that possibly blowing up wrong programs is actually the only effect.

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Comments

[R. Martinho Fernandes]

Yes.⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣⁣

[Nikos C.]

It's not a double check though. It's the only check.

[aiao]

@Nikos good point;), fixed

[KAlO2]

hey, override is NOT a keyword, it's kind of a grammar sugar. int override=42; // OK

[mots_g]

It additionally improves readability explaining the declared function is overridden ;)

[SasQ]

@KAlO2: This makes me wonder what would happen if I define int override; as a data member of my class ;-J Wouldn't this "not-a-keyword" conflict with that member then?

[SasQ]

Or better: a member function called "override" :) e.g. int override(float override) override {...}. What a mess... :D

[Cinch]

So, uh... When will C++11 become standard enough that they start teaching things like this at my local 4-year? When will they know?!

[underscore_d]

@SasQ Of course override as a regular identifier cannot conflict with the override specifier, since the latter can only appear in one location, wherein the former can never be syntactically valid. That's precisely why they set it up this way, to enable override and final to have special meanings in the single context where they matter, while not divesting users of the previously available ability to use them as normal identifiers.

[Lightness Races in Orbit]

+1: Though, unfortunately, it's a bit of a red herring when people suggest that the new override feature "fixes" this; you have to remember to use it, just as you should have remembered to write the const ;)

[aschepler]

I just realized explicit class definitions didn't make it into C++11. Huh.

[Christian Rau]

@aschepler And what would an explicit class definition do? Never heard about that at all.

[legends2k]

@LightnessRacesinOrbit: Yes, it's not fool proof; however, remembering a general rule (of madly writing override when one intends to do it) is more likely than remembering corner cases i.e. there's no generality in copying functions of different prototypes, only irregularities like missing const or writing char instead of int, etc.

[Lightness Races in Orbit]

@legends2k: I suppose.

[Kerrek SB]

@LightnessRacesinOrbit: As much as I agree that most defaults in C++ are the wrong way round, I don't really see how you would change the "defaultness" of override. Apply it to every function? That'd be nonsense. (And I think non-virtual by default is one of the few things C++ got right.) Apply it to all virtual functions? That'd also get old really fast in all your base classes. A common style now is to use precisely one of virtual, override and final, whichever is strongest, and I'm not sure there's room for different defaults. Very much unlike with const.

[Lightness Races in Orbit]

@KerrekSB: I never suggested doing either of those things.

[Kerrek SB]

@LightnessRacesinOrbit: No, of course, I just wanted to make it clear that even though you still have to remember to use override, it's not necessarily the case that a better "fix" was missed, I think.

[Lightness Races in Orbit]

@KerrekSB: Indeed.

[iammilind]

@Light, the best use case of override specifier is mentioned in this answer, which is more futuristic than immediate. The answer suggests that, keep the override with the virtual method. In future when one mistakenly changes the signature, its usefulness kicks in.

[Lightness Races in Orbit]

@iammilind: Agreed.

[Disillusioned]

+1. Though while override is a great way to discover such problems, good unit test coverage should also help.

[Cheers and hth. - Alf]

Using virtual in a declaration that also has override is counter-productive. It's more to write, and more to read. stackoverflow.com/a/39932616/464581

[parsley72]

Core guideline C.128 states "Virtual functions should specify exactly one of virtual, override, or final." github.com/isocpp/CppCoreGuidelines/blob/master/…

[Wolf]

That's exactly why I'm so excited about that new specifier. The only problem is that this feature has to be already applied as to prevent errors caused by changes in base classes. ;-)