I would like to enable C4263 (Visual C++) warning on our code base, however, the warning gives out some false positives. We would like to disable the warning such that only the false positives disappear. I tried to simplify the issue with some generic code:
class B {
public:
virtual void funcA();
virtual void funcB();
};
class D : public B
{
virtual void funcA(int);
virtual void funcB(int);
};
With this code we I get following warnings:
void D::funcA(int)' : member function does not override any base class virtual member function
void D::funcB(int)' : member function does not override any base class virtual member function
What I am trying to achieve is to disable the warning for funcA (or funcB) and let the rest of the class be affected by it.
I tried putting
#pragma warning(push)
#pragma warning(disable:4263)
.
.
.
#pragma warning(pop)
around funcA but that does not solve the problem. If the pragmas wrap the whole class that both of the warnings disappear.
Any ideas?
This is a strange error given the documentation for it:
https://msdn.microsoft.com/en-us/library/ay4h0tc9.aspx?f=255&MSPPError=-2147217396
'function' : member function does not override any base class virtual
member function
A class function definition has the same name as a virtual function in
a base class but not the same number or type of arguments. This
effectively hides the virtual function in the base class.
The last sentence is the interesting one, and it suggested to me that if you unhide the base class functions, the warning will no longer appear.
And indeed this is the case. The following code does not output C4263:
class B {
public:
virtual void funcA();
virtual void funcB();
};
class D : public B
{
using B::funcA;
using B::funcB;
virtual void funcA(int);
virtual void funcB(int);
};
The warning seems a little strange. If you're dispatching from the base class pointer, it doesn't matter what functions the derived class hides, as they won't be hidden when using a base class pointer. But herein lies the answer!
What's actually happening here is the compiler is guessing your intentions. Because you are introducing a new signature, it means you will be using either polymorphic or non-polymorphic dispatch using a derived pointer (not a base pointer). If you were not doing this, it would be impossible to call your overload. The compiler figures that if you are doing this, you will be hiding the un-overridden functions. And that's what the warning is about.
In an example:
struct Base
{
virtual void DoThing(int)
{
std::cout << "INT " << std::endl;
}
};
struct Derived: public Base
{
virtual void DoThing(char) // Add a function to handle chars
{
std::cout << "CHAR " << std::endl;
}
};
int main()
{
Derived *derived = new Derived;
Base *base = derived;
base->DoThing(1);
derived->DoThing(1);
derived->DoThing('a');
}
This outputs:
INT CHAR CHAR
The intention may have been to add an overload to handle a different case, but instead it's hiding all the existing overloads. The warning is correct given the rules of the language. The warning isn't exact, it's trivial to determine cases where it doesn't get invoked but it should. It in-fact does the opposite of false-warnings :)
To defeat this warning, the using declaration should be used.
Related
Here is the problem: I keep getting the unimplemented pure virtual method error when trying to compile. I have implemented all of the pure virtual methods in the abstract base class. Any ideas?
here is the abstract base class:
class record{
public:
virtual int getID()=0;
virtual record *clone();
};
and the implementation:
class sdata: public record{
public:
sdata(std::string s = ""){data=s; ID=atoi(data.substr(0,8).c_str());}
virtual int getID(){return ID;}
private:
std::string data;
int ID;
};
sorry, here is the complete error message:
Unimplemented pure virtual method 'getID' in 'record'
Perhaps this bit of code is causing the error then:
int hashTable::hash(record *x) {
return floor(m * (x->getID() * A - floor(x->getID() * A)));
}
Without seeing the code causing the error, it's difficult to know exactly what's going on. If this is a compile-time error, I don't see anything here that would cause it.
However, if you're seeing a runtime error, the two most common causes of this that I can think of are:
(1) Calling the member function from within the base class's constructor or destructor (even indirectly).
(2) The derived class calling the base class's version of the function without it being implemented.
An example showing both of these errors would be:
struct Base {
Base()
{
call_foo(); // Oops, indirectly calls Base::foo() (Scenario 1)
}
void call_foo() const {
foo();
}
protected:
virtual void foo() const = 0;
};
struct Derived : Base {
protected:
virtual void foo() const {
Base::foo(); // Oops, unimplemented virtual base function (Scenario 2)
}
};
int main() {
Derived().call_foo();
}
== UPDATE: Possible compile-time error ==
I observe in your example code that record has a non-pure-virtual clone() member function returning a record *. Since record is abstract, you can't create a record directly (only its concrete subclasses). This suggests that your clone() member function should probably also be pure virtual; if it tries to (for example) return new record(), you will get an error that your base class has pure virtual functions.
It sounds like you have not implemented all the functions from the abstract base class. If a function in your base has the signature:
void SomeFuction() const;
And you implement the following in your derived class:
void SomeFuction();
Then you have have not implemented the function because you omitted the const. GCC should tell you what you did not implement.
Related: if you had a concrete implementation of the base, you would have hidden its name in the derived class. To find functions you are [accidentally] hiding, use -Woverloaded-virtual.
Suppose we have the following code:
class Base
{
public:
virtual void foo() const
{ cout << "In Base::foo\n"; }
}
class Derived : public Base
{public:
virtual void foo()
{ cout << "In Derived::foo\n"; }
}
void main()
{
Base* b = new Derived();
b->foo();
delete b;
}
It will give us the following output: In Base::foo.
Now suppose I want to get - without changing the main function - the follwing output instead the one given above:
In Derived::foo
As far as I understand, I should override the function foo() of in the base, to get the output of the method
foo() in the inheriting class which is class 'Derived'.
But the problem is that in that case I can't using the command override, becuase the method in the base is defined as constant , which in the other class it is not.
So, how should I if then overriding that method ?
In order to override the function void foo() const of the base class, you must declare the function void foo() const in the derived class. You've declared the function void foo() instead (note the lack of const), which doesn't override the function in the base, because it has a different declaration.
So, how should I if then overriding that method ?
Add the missing const qualifier.
P.S. Few other bugs in your program:
main must return int.
Class definitions must end in a semicolon.
Deleting a derived object through a pointer to base has undefined behaviour unless the destructor of the base is virtual.
I'm looking at some C++ example code, which effectively has the following:
class Foo : public mynamespace::Bar
{
public:
Foo()
{
// Do some stuff
}
void Delta() override
{
// Do some stuff
Bar::Delta();
}
};
I am having trouble understanding why the line Bar::Delta(); exists. Given that class Foo inherits class Bar, then surely when Foo::Delta() is called, this overrides anything existing in Bar::Delta(), and hence this line is redundant.
Or am I misunderstanding this whole inheritance thing? Maybe override doesn't override everything?
Bar::Delta();
is a function call. Isn't that what it looks like?
It's calling the base class version of Delta - by explicitly qualifying it as Bar::Delta - in addition to whatever extra stuff Foo's version does.
Maybe override doesn't override everything?
The override keyword just asks the compiler to verify you're really overriding a virtual function - otherwise, it's easy to accidentally write a new function which doesn't override anything (eg. because a parameter type is slightly different, or one version is const-qualified, or you changed the base class and forgot to update the derived class, or ...).
Here you are overriding the virtual function. That doesn't stop the base class implementation from existing, and as you've seen you can still call it.
You can (and should) test your intuition about this sort of thing yourself. Consider the trivial test code below. What do you think it will do?
Now actually run it. Were you right? If not, which part was unexpected?
#include <iostream>
using namespace std;
struct Bar {
virtual void Delta() {
cout << "Bar::Delta\n";
}
};
struct Foo : public Bar {
void Delta() override
{
cout << "Foo::Delta\n";
Bar::Delta();
}
};
int main()
{
cout << "b.Delta()\n";
Bar b;
b.Delta();
cout << "f.Delta()\n";
Foo f;
f.Delta();
cout << "pb->Delta()\n";
Bar *pb = &b;
pb->Delta();
cout << "pb->Delta()\n";
pb = &f;
pb->Delta();
}
This is a common pattern. In fact, the :: syntax for calling an overridden member function is there specifically for this situation.
It is very common for member function in a base class to perform some computation or action which can be done independently of the derived class, and let the derived class do things specific to the derivation.
Here is a fictitious example:
class Stock {
protected:
double totalDividend;
double baseDividend;
double adjustmentFactor;
public:
Stock(double d, double a)
: baseDividend(d), totalDividend(d), adjustmentFactor(a) {
}
virtual void double ComputeDividend() {
return totalDividend * adjustmentFactor;
}
};
class SpecialStock {
private:
double specialDividend;
public:
SpecialStock(double d, double sd, double a)
: Stock(d, a), specialDividend(sd) {
}
virtual void double ComputeDividend() override {
// Do some preparations
totalDividend = baseDividend + specialDividend;
// Call the overridden function from the base class
return Stock::ComputeDividend();
}
};
You seem to misunderstand what override means in c++. You should not really bother by it for your case. It's like a normal virtual function. override keyword is just a safety mechanism to ensure that a base class has matching function. It doesn't change anything other than semantic compile-time checks.
It is somewhat useful to guard against typical mismatches, such a const vs non-const versions, etc.
The virtual method mechanism does not replace the original member functions. It still is there in the Derived object. What happens is that a level of indirection is introduced, so a call to base.foo() uses a function pointer to call correct implementation which is Derived::Foo() in your case. But Base::Foo() still exists, and this is the syntax to "access" it. You can look up how exactly it work by searching materials on virtual methods.
Maybe override doesn't override everything?
The override specifier can be used to make compile-time checks when overriding a function that the function being overridden is virtual and is in fact being overridden.
why the line Bar::Delta(); exists
This line calls the base Delta function which might have some useful tasks to perform even though you have overridden it.
A simple example:
class Base
{
public:
virtual ~Base() {}
virtual void Foo()
{
// run tasks that are common to all derived types
}
};
class Derived : public Base
{
public:
void Foo() override
{
Base::Foo(); // we have to call this explicitly
// run tasks specific to the derived type
}
};
Delta in foo is a virtual method which "does some stuff" and then calls the base class's implementation of Delta.
override does not override anything, declaring the method as virtual does that. override is just for the compiler to throw syntax error in case the parent class doesn't have the Delta method.
What is the purpose of the final keyword in C++11 for functions? I understand it prevents function overriding by derived classes, but if this is the case, then isn't it enough to declare as non-virtual your final functions? Is there another thing I'm missing here?
What you are missing, as idljarn already mentioned in a comment is that if you are overriding a function from a base class, then you cannot possibly mark it as non-virtual:
struct base {
virtual void f();
};
struct derived : base {
void f() final; // virtual as it overrides base::f
};
struct mostderived : derived {
//void f(); // error: cannot override!
};
It is to prevent a class from being inherited. From Wikipedia:
C++11 also adds the ability to prevent inheriting from classes or simply preventing overriding methods in derived classes. This is done with the special identifier final. For example:
struct Base1 final { };
struct Derived1 : Base1 { }; // ill-formed because the class Base1
// has been marked final
It is also used to mark a virtual function so as to prevent it from being overridden in the derived classes:
struct Base2 {
virtual void f() final;
};
struct Derived2 : Base2 {
void f(); // ill-formed because the virtual function Base2::f has
// been marked final
};
Wikipedia further makes an interesting point:
Note that neither override nor final are language keywords. They are technically identifiers; they only gain special meaning when used in those specific contexts. In any other location, they can be valid identifiers.
That means, the following is allowed:
int const final = 0; // ok
int const override = 1; // ok
"final" also allows a compiler optimization to bypass the indirect call:
class IAbstract
{
public:
virtual void DoSomething() = 0;
};
class CDerived : public IAbstract
{
void DoSomething() final { m_x = 1 ; }
void Blah( void ) { DoSomething(); }
};
with "final", the compiler can call CDerived::DoSomething() directly from within Blah(), or even inline. Without it, it has to generate an indirect call inside of Blah() because Blah() could be called inside a derived class which has overridden DoSomething().
Nothing to add to the semantic aspects of "final".
But I'd like to add to chris green's comment that "final" might become a very important compiler optimization technique in the not so distant future. Not only in the simple case he mentioned, but also for more complex real-world class hierarchies which can be "closed" by "final", thus allowing compilers to generate more efficient dispatching code than with the usual vtable approach.
One key disadvantage of vtables is that for any such virtual object (assuming 64-bits on a typical Intel CPU) the pointer alone eats up 25% (8 of 64 bytes) of a cache line. In the kind of applications I enjoy to write, this hurts very badly. (And from my experience it is the #1 argument against C++ from a purist performance point of view, i.e. by C programmers.)
In applications which require extreme performance, which is not so unusual for C++, this might indeed become awesome, not requiring to workaround this problem manually in C style or weird Template juggling.
This technique is known as Devirtualization. A term worth remembering. :-)
There is a great recent speech by Andrei Alexandrescu which pretty well explains how you can workaround such situations today and how "final" might be part of solving similar cases "automatically" in the future (discussed with listeners):
http://channel9.msdn.com/Events/GoingNative/2013/Writing-Quick-Code-in-Cpp-Quickly
Final cannot be applied to non-virtual functions.
error: only virtual member functions can be marked 'final'
It wouldn't be very meaningful to be able to mark a non-virtual method as 'final'. Given
struct A { void foo(); };
struct B : public A { void foo(); };
A * a = new B;
a -> foo(); // this will call A :: foo anyway, regardless of whether there is a B::foo
a->foo() will always call A::foo.
But, if A::foo was virtual, then B::foo would override it. This might be undesirable, and hence it would make sense to make the virtual function final.
The question is though, why allow final on virtual functions. If you have a deep hierarchy:
struct A { virtual void foo(); };
struct B : public A { virtual void foo(); };
struct C : public B { virtual void foo() final; };
struct D : public C { /* cannot override foo */ };
Then the final puts a 'floor' on how much overriding can be done. Other classes can extend A and B and override their foo, but it a class extends C then it is not allowed.
So it probably doesn't make sense to make the 'top-level' foo final, but it might make sense lower down.
(I think though, there is room to extend the words final and override to non-virtual members. They would have a different meaning though.)
A use-case for the 'final' keyword that I am fond of is as follows:
// This pure abstract interface creates a way
// for unit test suites to stub-out Foo objects
class FooInterface
{
public:
virtual void DoSomething() = 0;
private:
virtual void DoSomethingImpl() = 0;
};
// Implement Non-Virtual Interface Pattern in FooBase using final
// (Alternatively implement the Template Pattern in FooBase using final)
class FooBase : public FooInterface
{
public:
virtual void DoSomething() final { DoFirst(); DoSomethingImpl(); DoLast(); }
private:
virtual void DoSomethingImpl() { /* left for derived classes to customize */ }
void DoFirst(); // no derived customization allowed here
void DoLast(); // no derived customization allowed here either
};
// Feel secure knowing that unit test suites can stub you out at the FooInterface level
// if necessary
// Feel doubly secure knowing that your children cannot violate your Template Pattern
// When DoSomething is called from a FooBase * you know without a doubt that
// DoFirst will execute before DoSomethingImpl, and DoLast will execute after.
class FooDerived : public FooBase
{
private:
virtual void DoSomethingImpl() {/* customize DoSomething at this location */}
};
final adds an explicit intent to not have your function overridden, and will cause a compiler error should this be violated:
struct A {
virtual int foo(); // #1
};
struct B : A {
int foo();
};
As the code stands, it compiles, and B::foo overrides A::foo. B::foo is also virtual, by the way. However, if we change #1 to virtual int foo() final, then this is a compiler error, and we are not allowed to override A::foo any further in derived classes.
Note that this does not allow us to "reopen" a new hierarchy, i.e. there's no way to make B::foo a new, unrelated function that can be independently at the head of a new virtual hierarchy. Once a function is final, it can never be declared again in any derived class.
The final keyword allows you to declare a virtual method, override it N times, and then mandate that 'this can no longer be overridden'. It would be useful in restricting use of your derived class, so that you can say "I know my super class lets you override this, but if you want to derive from me, you can't!".
struct Foo
{
virtual void DoStuff();
}
struct Bar : public Foo
{
void DoStuff() final;
}
struct Babar : public Bar
{
void DoStuff(); // error!
}
As other posters pointed out, it cannot be applied to non-virtual functions.
One purpose of the final keyword is to prevent accidental overriding of a method. In my example, DoStuff() may have been a helper function that the derived class simply needs to rename to get correct behavior. Without final, the error would not be discovered until testing.
Final keyword in C++ when added to a function, prevents it from being overridden by derived classes.
Also when added to a class prevents inheritance of any type.
Consider the following example which shows use of final specifier. This program fails in compilation.
#include <iostream>
using namespace std;
class Base
{
public:
virtual void myfun() final
{
cout << "myfun() in Base";
}
};
class Derived : public Base
{
void myfun()
{
cout << "myfun() in Derived\n";
}
};
int main()
{
Derived d;
Base &b = d;
b.myfun();
return 0;
}
Also:
#include <iostream>
class Base final
{
};
class Derived : public Base
{
};
int main()
{
Derived d;
return 0;
}
Final keyword have the following purposes in C++
If you make a virtual method in base class as final, it cannot be overridden in the derived class. It will show a compilation error:
class Base {
public:
virtual void display() final {
cout << "from base" << endl;
}
};
class Child : public Base {
public:
void display() {
cout << "from child" << endl;
}
};
int main() {
Base *b = new Child();
b->display();
cin.get();
return 0;
}
If we make a class as final, it cannot be inherited by its child classes:
class Base final {
public:
void displayBase() {
cout << "from base" << endl;
}
};
class Child :public Base {
public:
void displayChild() {
cout << "from child" << endl;
}
};
Note: the main difference with final keyword in Java is ,
a) final is not actually a keyword in C++.
you can have a variable named as final in C++
b) In Java, final keyword is always added before the class keyword.
Supplement to Mario Knezović 's answer:
class IA
{
public:
virtual int getNum() const = 0;
};
class BaseA : public IA
{
public:
inline virtual int getNum() const final {return ...};
};
class ImplA : public BaseA {...};
IA* pa = ...;
...
ImplA* impla = static_cast<ImplA*>(pa);
//the following line should cause compiler to use the inlined function BaseA::getNum(),
//instead of dynamic binding (via vtable or something).
//any class/subclass of BaseA will benefit from it
int n = impla->getNum();
The above code shows the theory, but not actually tested on real compilers. Much appreciated if anyone paste a disassembled output.
I would like a class B not to be able to redefine one of the member function of its base class A. Is there a way to do that?
EDIT:
Thanks for the answers. Can I prevent non-virtual member functions from being overridden as well?
If you mean disallowing a derived class to override a virtual function of its base class, C++11 introduced a way to do that with a final virt-specifier (as it's called in the standard):
struct B{
virtual void f() final;
};
struct D : B{
void f(); // error: 'B::f' marked 'final'
};
I don't know which compilers support that, though. Probably GCC 4.7 and Clang.
If your methods are virtual, in C++11 you can prevent overriding them with final.
class A
{
public:
virtual void foo() final;
};
class B : public A
{
public:
void foo(); // <-- error
};
You can't prevent hiding.
Can I prevent non-virtual member functions from being overridden as
well?
No you can't, BUT from another side you can... if you're not shore if will you acctualy override base method or not in your derived class you can check whther you're overriding an method or not using a keywod override
when applying that keywod to your method you make shore that you accutaly realy want override that mathod.
so if signature of inherted method is the same as the signature of base method you'll get compile time error, say that derived method does not override base method.
so using override keyword you are telling the compiler that you DO NOT want to everride it.
by not using override keywod you never know wether you override it or not.
consider following example:
class A
{
public:
void f() {std::cout << "A";}
};
class B : public A
{
public:
void f() override {std::cout << "B";}
};
this result in compile time error because you can't override base method.
by not using the override keyword you'll of course be able to override it.
override keywod is introduced in C++0x standard and is supported by visual studio 2010.