C++ : Automatically run function when derived class is constructed - c++

So I recently accidentally called some virtual functions from the constructor of a base class, i.e. Calling virtual functions inside constructors.
I realise that I should not do this because overrides of the virtual function will not be called, but how can I achieve some similar functionality? My use-case is that I want a particular function to be run whenever an object is constructed, and I don't want people who write derived classes to have to worry about what this is doing (because of course they could call this thing in their derived class constructor). But, the function that needs to be called in-turn happens to call a virtual function, which I want to allow the derived class the ability to override if they want.
But because a virtual function gets called, I can't just stick this function in the constructor of the base class and have it get run automatically that way. So I seem to be stuck.
Is there some other way to achieve what I want?
edit: I happen to be using the CRTP to access other methods in the derived class from the base class, can I perhaps use that instead of virtual functions in the constructor? Or is much the same issue present then? I guess perhaps it can work if the function being called is static?
edit2: Also just found this similar question: Call virtual method immediately after construction

If really needed, and you have access to the factory.
You may do something like:
template <typename Derived, typename ... Args>
std::unique_ptr<Derived> Make(Args&&... args)
{
auto derived = std::make_unique<Derived>(std::forward<Args>(args));
derived->init(); // virtual call
return derived;
}

There is no simple way to do this. One option would be to use so-called virtual constructor idiom, hide all constructors of the base class, and instead expose static 'create' - which will dynamically create an object, call your virtual override on it and return (smart)pointer.
This is ugly, and what is more important, constrains you to dynamically created objects, which is not the best thing.
However, the best solution is to use as little of OOP as possible. C++ strength (contrary to popular belief) is in it's non-OOP specific traits. Think about it - the only family of polymorphic classess inside standard library are streams, which everybody hate (because they are polymorphic!)

I want a particular function to be run whenever an object is constructed, [... it] in-turn happens to call a virtual function, which I want to allow the derived class the ability to override if they want.
This can be easily done if you're willing to live with two restrictions:
the constructors in the entire class hierarchy must be non-public, and thus
a factory template class must be used to construct the derived class.
Here, the "particular function" is Base::check, and the virtual function is Base::method.
First, we establish the base class. It has to fulfill only two requirements:
It must befriend MakeBase, its checker class. I assume that you want the Base::check method to be private and only usable by the factory. If it's public, you won't need MakeBase, of course.
The constructor must be protected.
https://github.com/KubaO/stackoverflown/tree/master/questions/imbue-constructor-35658459
#include <iostream>
#include <utility>
#include <type_traits>
using namespace std;
class Base {
friend class MakeBase;
void check() {
cout << "check()" << endl;
method();
}
protected:
Base() { cout << "Base()" << endl; }
public:
virtual ~Base() {}
virtual void method() {}
};
The templated CRTP factory derives from a base class that's friends with Base and thus has access to the private checker method; it also has access to the protected constructors in order to construct any of the derived classes.
class MakeBase {
protected:
static void check(Base * b) { b->check(); }
};
The factory class can issue a readable compile-time error message if you inadvertently use it on a class not derived from Base:
template <class C> class Make : public C, MakeBase {
public:
template <typename... Args> Make(Args&&... args) : C(std::forward<Args>(args)...) {
static_assert(std::is_base_of<Base, C>::value,
"Make requires a class derived from Base");
check(this);
}
};
The derived classes must have a protected constructor:
class Derived : public Base {
int a;
protected:
Derived(int a) : a(a) { cout << "Derived() " << endl; }
void method() override { cout << ">" << a << "<" << endl; }
};
int main()
{
Make<Derived> d(3);
}
Output:
Base()
Derived()
check()
>3<

If you take a look at how others solved this problem, you will notice that they simply transferred the responsibility of calling the initialization function to client. Take MFC’s CWnd, for instance: you have the constructor and you have Create, a virtual function that you must call to have a proper CWnd instantiation: “these are my rules: construct, then initialize; obey, or you’ll get in trouble”.
Yes, it is error prone, but it is better than the alternative: “It has been suggested that this rule is an implementation artifact. It is not so. In fact, it would be noticeably easier to implement the unsafe rule of calling virtual functions from constructors exactly as from other functions. However, that would imply that no virtual function could be written to rely on invariants established by base classes. That would be a terrible mess.” - Stroustrup. What he meant, I reckon, is that it would be easier to set the virtual table pointer to point to the VT of derived class instead of keep changing it to the VT of current class as your constructor call goes from base down.
I realise that I should not do this because overrides of the virtual function will not be called,...
Assuming that the call to a virtual function would work the way you want, you shouldn't do this because of the invariants.
class B // written by you
{
public:
B() { f(); }
virtual void f() {}
};
class D : public B // written by client
{
int* p;
public:
D() : p( new int ) {}
void f() override { *p = 10; } // relies on correct initialization of p
};
int main()
{
D d;
return 0;
}
What if it would be possible to call D::f from B via VT of D? You will use an uninitialized pointer, which will most likely result in a crash.
...but how can I achieve some similar functionality?
If you are willing to break the rules, I guess that it might be possible to get the address of desired virtual table and call the virtual function from constructor.

Seems you want this, or need more details.
class B
{
void templateMethod()
{
foo();
bar();
}
virtual void foo() = 0;
virtual void bar() = 0;
};
class D : public B
{
public:
D()
{
templateMethod();
}
virtual void foo()
{
cout << "D::foo()";
}
virtual void bar()
{
cout << "D::bar()";
}
};

Related

What are the advantages of having a definition of pure virtual method In C++? [duplicate]

I did a simple test today:
struct C{virtual void f()=0;};
void C::f(){printf("weird\n");}
The program is OK, but is weird to me, when we use =0 it means the function body should be defined in the inherited classes, but it seems I can still give it implementation function.
I tried both GCC and VC, both OK. So it seems to me this should be part of C++ standard.
But why this is not a syntax error?
A reason I could think of is like C# having both 'interface' and 'abstract' keywords, interface can't have an implementation, while abstract could have some implementations.
Is this the case for my confusion, that C++ should support such a kind of weird syntax?
C++ Supports pure virtual functions with an implementation so class designers can force derived classes to override the function to add specific details , but still provide a useful default implementation that they can use as a common base.
Classic example:
class PersonBase
{
private:
string name;
public:
PersonBase(string nameIn) : name(nameIn) {}
virtual void printDetails() = 0
{
std::cout << "Person name " << name << endl;
}
};
class Student : public PersonBase
{
private:
int studentId;
public:
Student(string nameIn, int idIn) : PersonBase(nameIn), studentId(idIn) { }
virtual void printDetails()
{
PersonBase::printDetails(); // call base class function to prevent duplication
std::cout << "StudentID " << studentId << endl;
}
};
Others mentioned language consistency with the destructor, so I'll go for a software engineering stand-point:
It's because the class you are defining may have a valid default implementation, but calling it is risky/expansive/whatever. If you don't define it as pure virtual, derived classes will inherit that implementation implicitly. And may never know until run-time.
If you define it as pure virtual, a derived class must implement the function. And if it's okay with the risk/cost/whatever, it can call the default implementation statically as Base::f();
What's important is that it's a conscious decision, and the call is explicit.
Basically, the best of both worlds (or the worst...).
The derived class is required to implement the pure virtual method, the designer of the base class requires this for some reason. And the base class also provides a default implementation of this method that, if the derived class desires or requires it, can be used.
So some sample code could look like;
class Base {
public:
virtual int f() = 0;
};
int Base::f() {
return 42;
}
class Derived : public Base {
public:
int f() override {
return Base::f() * 2;
}
};
So what is a common use case...
A common use case for this technique is related to the destructor - basically the designer of the base class desires that it is an abstract class, but none of the methods make much sense as being pure virtual functions. The destructor is a feasible candidate.
class Base {
public:
~Base() = 0;
};
Base::~Base() { /* destruction... */ }
A pure virtual function must be overriden in subclasses. However, you can provide a default-implementation, that will work for sub-classes, but might not be optimal.
A constructed use case is for abstract shapes, e.g.
class Shape {
public:
virtual Shape() {}
virtual bool contains(int x, int y) const = 0;
virtual int width() const = 0;
virtual int height() const = 0;
virtual int area() const = 0;
}
int Shape::area() const {
int a = 0;
for (int x = 0; x < width(); ++x) {
for (int y = 0; y < height(); ++y) {
if (contains(x,y)) a++;
}
}
return a;
}
The area method will work for any shape, but is highly inefficient. Subclassers are encouraged to provide a suitable implementation, but if there is none available, they still can explicitely call the base class's method
Pure virtual means "child must override".
So:
struct A{ virtual void foo(){}; };
struct B:A{ virtual void foo()=0; };
struct C:B{ virtual void foo(){}; };
struct D:C{ virtual void foo()=0; };
void D::foo(){};
struct E:D{ virtual void foo(){D::foo();}; };
A has a virtual foo.
B makes it abstract. Before making an instance, derived types must implement it now.
C implements it.
D makes it abstract, and adds an imllementation.
E implements it by calling D's implementation.
A, C and E can have instances created. B and D cannot.
The technique of abstract with implementation can be used to provide a partial or inefficient implementation that derived types can call explicitly when they want to use it, but do not get "by default" because that would be ill advised.
Another intersting use case is where the parent interface is in flux, and tue code base is large. It has a fully functional implementation. Children who use the default must repeat the signature and forward explicitly to it. Those that want to override simply override.
When the base class sigrnature changes, the code will fail to compile unless every child either explicitly calls the default or properly overrides. Prior to the override keyword this was the only way to ensure you did not accidentally create a new virtual function instead of overriding a parent, and it remains the only way where the policy is enforced in the parent type.
Please note that you cannot instantiate an object with pure virtual methods.
Try to instantiate:
C c;
with VC2015, there is an error as expected:
1>f:\dev\src\consoleapplication1\consoleapplication1.cpp(12): error C2259: 'C': cannot instantiate abstract class
1>f:\dev\src\consoleapplication1\consoleapplication1.cpp(12): note: due to following members:
1>f:\dev\src\consoleapplication1\consoleapplication1.cpp(12): note: 'void C::f(void)': is abstract
1>f:\dev\src\consoleapplication1\consoleapplication1.cpp(6): note: see declaration of 'C::f'
To answer your question:
The mechanisms only declares the function to be pure virtual, but there is still the virtual function table and the baseclass. It will avoid you instanciate Baseclass (C), but does not avoid using it:
struct D : public C { virtual void f(); };
void D::f() { printf("Baseclass C::f(): "); C::f(); }
...
D d;
d.f();
The destructor must be defined, even if it is pure virtual. If you don't define the destructor the compiler will generate one.
Edit: you can't leave destructor declared without define, will cause link error.
You can anyway call the body of the function from derived classes.
You can implement the body of a pure virtual function to provide a default behavior, and at the same time you want that the designer of the derived class use that function explicitly.

Private overriden virtual functions in derived class

Is there any point to making virtual member functions, overridden from a base class private, if those are public in the base class?
struct base {
virtual void a();
};
struct derived : base {
// ...
private:
void a() override;
};
If you are forced to do a 2-phase construction on the implementation class (i.e. have an init() method as well as or instead of a constructor that has to be called (I know, but there are reasons), then this stops you calling any /other/ methods directly on the instance pointer before you pass it back as an interface pointer. Go the extra mile, make the inheritance private, and have your one public init function return the interface pointer!
Another reason is you just don't /need/ to write public: in a final implementation class declaration, so then by default everything is private. But why you would do that and use struct instead of class I don't know. Perhaps this was converted from class at some point due to a style war?
Looking at your design, I see one cannot call derived::a directly, but only through a base interface.
Is there any point? Consider that, once we have a derived instance, we can always up-cast to its base, so given
derived d;
while d.a() wouldn't compile, we can always do
base & b = d;
b.a(); //which actually calls derived::a
In other words: derived::a is not that private, after all, and I would discourage this design, which can be confusing to the user.
Things change if the members private in derived are private in base, as well: this time it is clear that they just cannot be called directly, outside base or derived.
Let's say we have a couple of functions, and want them to be called conditionally, according to a value passed as an argument to a third one:
struct base
{
void dosomething(bool x)
{
if(x)
{
do_this();
}
else
{
do_that();
}
}
private:
virtual void do_this(){}
virtual void do_that(){}
};
Thus a derived class could be like:
struct derived : base
{
private:
void do_this() override { }
void do_that() override { }
};
and no other class can call them, unless it extended base itself:
derived d;
d.dosomething(true); //will call do_this() in derived
d.dosomething(false); //will call do_that() in derived
d.do_that() //won't compile
Yes, if you inherit the base class as private. Otherwise, it is more of a weird explicit-like restriction - user has to has to make an explicit conversion to use the function - it is generally ill advised as few will be able to comprehend the author's intention.
If you want to restrict some functions from base class, make a private/protected inheritance and via using keyword declare which base-methods you want to be protected/public in the derived class.
The same reasoning as for non-virtual methods applies: If only the class itself is supposed to call it make it private.
Consider the template method pattern:
struct base {
void foo() { a() ; b(); }
virtual void a() = 0;
virtual void b() = 0;
};
struct derived : base {
private:
void a() override {}
void b() override {}
};
int main()
{
derived().foo();
}
Perhaps a and b should have been protected, but anyhow the derived can change accesibility and it requires some documentation so that derived knows how it is supposed to implement a and b.

Inheritance with an override function

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.

C++ hidding of member functions in inheritance hierarchy staring with CRTP

Yesterday, I wrote some code and i would really appreciate a judgement if this is good or bad practice. And if its bad, what could go wrong.
The construct is as followed:
The base class A comes from an API sadly as a template. The goal was to be able to put derived classes from A in a std::vector. I achieved this with the base class B from which all derived classes that inherit from A will inherit. Without the pure-virtual function in B the foo() function from A was called.
With the pure-virtual function in B the foo() version from C get called. This is exactly what I want.
So is this bad practice?
EDIT:
To clear some misunderstandings in my example code out:
template <class T>
class A
{
public:
/* ctor & dtor & ... */
T& getDerived() { return *static_cast<T*>(this); }
void bar()
{
getDerived().foo(); // say whatever derived class T will say
foo(); // say chicken
}
void foo() { std::cout << "and chicken" << std::endl; }
};
class B : public A<B>
{
public:
/* ctor & dtor */
virtual void foo() = 0; // pure-virtual
};
class C : public B
{
public:
/* ctor & dtor */
virtual void foo() { std::cout << "cow"; }
};
class D : public B
{
public:
/* ctor & dtor */
virtual void foo() { std::cout << "bull"; }
};
The std::vector shall contain both, C as well as D and
void main()
{
std::vector<B*> _cows;
_cows.push_back((B*)new C()));
_cows.push_back((B*)new D()));
for(std::vector<B*>::size_type i = 0; i != _cows.size(); ++i)
_cows[i].bar();
}
with output
cow and chicken
bull and chicken
is desired.
As far as I know is there no other way to store classes derived from a template class in a container than the use of a base class. If I use a base class for my derived classes, e.g., B i must cast every instance within the vector back to its proper class. But at the time bar() is called, I don't know the exact type.
I call it bad practice for doing possibly confusing and obfuscating things.
1) A function name should in all base and sub-classes either be virtual or non-virtual.
mixing overriding and overloading within the inheritance hierachry is a very bad idea.
No one really expects something like that.
Thus if A::foo should be virtual as well, or B and C foo should not.
2) Preferably base classes should be interfaces.
As such A:foo should be pure virtual and not B::foo.
2b) If a base class has already provided a default implementation don't declare the function pure virtual in a sub-class.

CRTP Dispatch in C++11

Say I have the following code:
template <class Derived>
class Base {
public:
virtual void foo_impl() = 0;
void foo() {
static_cast<Derived*>(this)->foo_impl(); //A
(*static_cast<Derived*>(this)).foo_impl(); //B
}
};
class Derived : public Base<Derived> {
private:
void foo_impl() {
bar();
}
};
A few questions:
Will line A generate a virtual function call? Although the majority of what I can find on the internet recommends doing things this way, to me I don't see how the compiler can do static dispatch considering that a pointer to Derived could still actually point to an object of type Derived2 where Derived2 : public Derived.
Does line B fix the issue I brought up in my previous point (if applicable)? It seems like it would, considering that now the call is not on a pointer any more and thus using *. would avoid a virtual function call. But if the compiler treats the dereferenced cast as a reference type, it could still generate a virtual function call... in that case, what is the workaround?
Does adding the C++11 final keyword to foo_impl() change how the compiler would act in either (or any other relevant) case?
Will line A generate a virtual function call?
Yes. foo_impl() is virtual and Derived overrides it. Even though foo_impl() in Derived is not explicitly tagged as virtual, it is in the base class, and this is enough to make it a virtual function.
Does line B fix the issue I brought up in my previous point (if applicable)?
No. It does not matter if the call is on a pointer or on a reference: the compiler still won't know whether you are invoking the function foo_impl() on an instance of a class that derives from Derived, or on a direct instance of Derived. Thus, the call is performed through a vtable.
To see what I mean:
#include <iostream>
using namespace std;
template <class Derived>
class Base {
public:
virtual void foo_impl() = 0;
void foo() {
static_cast<Derived*>(this)->foo_impl();
(*static_cast<Derived*>(this)).foo_impl();
}
};
class Derived : public Base<Derived> {
public:
void foo_impl() {
cout << "Derived::foo_impl()" << endl;
}
};
class MoreDerived : public Derived {
public:
void foo_impl() {
cout << "MoreDerived::foo_impl()" << endl;
}
};
int main()
{
MoreDerived d;
d.foo(); // Will output "MoreDerived::foo_impl()" twice
}
Finally:
Does adding the C++11 final keyword to foo_impl() change how the compiler would act in either (or any other relevant) case?
In theory, yes. The final keyword would make it impossible to override that function in subclasses of Derived. Thus, when performing a function call to foo_impl() through a pointer to Derived, the compiler could resolve the call statically. However, to the best of my knowledge, compilers are not required to do so by the C++ Standard.
CONCLUSION:
In any case, I believe what you actually want to do is not to declare the foo_impl() function at all in the base class. This is normally the case when you use the CRTP. Additionally, you will have to declare class Base<Derived> a friend of Derived if you want it to access Derived's private function foo_impl(). Otherwise, you can make foo_impl() public.
The common idiom for the CRTP does not involve declaring the pure virtual functions in the base. As you mention in one of the comments, that means that the compiler will not enforce the definition of the member in the derived type (other than through use, if there is any use of foo in the base, that requires the presence of foo_impl in the derived type).
While I would stick to the common idiom and not define the pure virtual function in the base, but, if you really feel you need to do it, you can disable dynamic dispatch by adding extra qualification:
template <class Derived>
class Base {
public:
virtual void foo_impl() = 0;
void foo() {
static_cast<Derived*>(this)->Derived::foo_impl();
// ^^^^^^^^^
}
};
The use of the extra qualification Derived:: disables dynamic dispatch, and that call will be statically resolved to Derived::foo_impl. Note that this comes will all of the usual caveats: you have a class with a virtual function and paying the cost of the virtual pointer per object, but you cannot override that virtual function in a most derived type, as the use in the CRTP base is blocking dynamic dispatch...
The extra verbiage in lines A and B have absolutely no effect on
the generated code. I don't know who recommends this (I've never seen
it), but in practice, the only time it might have an effect is
if the function isn't virtual. Just write foo_impl(), and be
done with it.
There is a means of avoiding the virtual function call if the
compiler knows the derived type. I've seen it used for
vector-like classes (where there are different implementations,
e.g. normal, sparse, etc. of the vector):
template <typename T>
class Base
{
private:
virtual T& getValue( int index ) = 0;
public:
T& operator[]( int index ) { return getValue( index ); }
};
template <typename T>
class Derived : public Base<T>
{
private:
virtual T& getValue( int index )
{
return operator[]( index );
}
public:
T& operator[]( index )
{
// find element and return it.
}
};
The idea here is that you normally only work through references
to the base class, but if performance becomes an issue, because
you're using [] in a tight loop, you can dynamic_cast to the
derived class before the loop, and use [] on the derived
class.