Let's say I'm trying to define an abstract base class for "mathematical functions" - i.e. things that look like y=f(x) where both x and y are numeric types, and x is in the "domain" of f. Here's what my abstract base class might look like:
template<typename T>
class AbstractFunction{
public:
virtual bool CheckDomain(T) const = 0;
virtual T operator()(T) const = 0;
};
This says that any concrete class derived from AbstractFunction must implement a CheckDomain method and an "evaluation operator" (). Now what I'd like to do is require that the evaluation operator calls the CheckDomain function before evaluating the function. Is there a neat way to do some sort of "partial implementation" for operator()(T)? Obviously this would mean it's no longer a pure virtual.
I thought I had it figured out with the following construction, but I must be missing something because if I create ConcreteFunction f and try to evaluate f(1.0), it doesn't use the derived class methods (it always returns 0 for CheckDomain(x)).
class AbstractFunction{
public:
virtual bool CheckDomain(double) const = 0;
virtual double EvalFunction(double) const = 0;
virtual double operator()(double);
};
double AbstractFunction::operator()(double x){
bool D = CheckDomain(x);
if(D==1){
return EvalFunction(x);
}
else{
std::runtime_error("Error: Out of Domain");
}
}
class ConcreteFunction : public AbstractFunction{
public:
bool CheckDomain(double x) const {(x>0.0)? 1:0;}
double EvalFunction(double x) const {return x*x;}
// Don't need to define operator(), should inherit from AbstractFunction.
};
I'm sure I'm missing something easy, any thoughts?
Actually, my method works! I just spaced out on my definition of ConcreteFunction::CheckDomain. I did the domain check, but didn't return it! The following code compiles and runs as desired:
#include<stdexcept>
#include<iostream>
class AbstractFunction{
public:
virtual bool CheckDomain(double) const = 0;
virtual double EvalFunction(double) const = 0;
virtual double operator()(double) const;
};
double AbstractFunction::operator()(double x) const{
bool D = CheckDomain(x);
if(D==1){
return EvalFunction(x);
}
else{
throw std::runtime_error("Error: Out of Domain");
}
}
class ConcreteFunction : public AbstractFunction{
public:
bool CheckDomain(double x) const {return (x>0.0)? 1:0;}
double EvalFunction(double x) const {return x*x;}
// Don't need to define operator(), should inherit from AbstractFunction.
};
int main()
{
ConcreteFunction f;
std::cout<<"f(1) = "<<f(2.0)<<std::endl; //OK, returns 4
std::cout<<"f(-1) = "<<f(-1.0)<<std::endl; //Throws runtime error, out of domain.
return 0;
}
Related
I have a base product class with a few private members and a public getter that derived classes inherit. I would like to disqualify instantiation, since the class is intended for use with an abstract factory. I thought protected con/destructors might work, however, this breaks my smart pointers. Friending seems like a useful disaster. Is there a well-known solution to this, or should I resign myself to the fact that any client who has the factory injected must also know enough to instantiate the base product?
class Product
{
private:
char type_name;
char size_name;
public:
Product(char, char);
virtual ~Product() {}
void Print();
};
Use a token key.
private:
Product(char, char);
struct key_t{explicit key_t(int){}};
static key_t key(){return key_t(0);}
public:
Product(key_t, char a, char b):Product(a,b){}
static std::shared_ptr<Product> make_shared(char a, char b){ return std::make_shared<Product>(key(),a,b); }
anyone with a Product::key_t can construct a Product without being a friend. And without the key, you cannot.
This lets Product pass creation-rights as a value.
Smart pointers with configurable destroy code can use similar techniques. But I'd just make the destructor public.
Your static member function, or friend function, which is the factory should have no problem with calling protected constructors and returning a smart pointer. Generally plan to return a std::unique_ptr<BaseClass> which can be converted into a std::shared_ptr if the caller wants that instead.
Make the virtual destructor public.
Update: Don't bother making the factory a friend. You only need to prevent the construction of the base and intermediate classes. Make them effectively hidden and private by hiding the implementation classes in their own source file. Or an anonymous namespace I suppose.
Here have some code of how I would do it:
#include <iostream>
#include <memory>
#include <string>
// ITest is the only class any other code file should ever see.
class ITest {
protected:
ITest() = default;
public:
virtual ~ITest() = 0;
virtual int getX() const = 0;
virtual int getY() const = 0;
};
// Destructors must always have an implementation even if they are pure virtual.
ITest::~ITest() {}
std::ostream &operator<<(std::ostream &os, const ITest &x) {
return os << '[' << x.getX() << ',' << x.getY() << ']';
}
// Declaration of constructTest factory function.
// Its definition should be hidden in a cpp file.
std::unique_ptr<ITest> constructTest(int x);
// The main function does not need to know anything except the ITest interface
// class and the constructTest function declaration.
int main(int argc, char *argv[]) {
int val = 0;
if (argc > 1)
val = std::stoi(argv[1]);
auto p = constructTest(val);
std::cout << *p << std::endl;
}
// These classes should be defined in a private header file or in a cpp file.
// Should not be visible to any other code. It has no business knowing.
// Hiding all of this implementation is sort of the point of abstract interface
// classes and factory function declarations.
class TestBase : public ITest {
private:
int x = 0;
int y = 0;
protected:
TestBase(int x = 0, int y = 0) : x(x), y(y){};
public:
int getX() const override { return x; }
int getY() const override { return y; }
};
class TestA final : public TestBase {
public:
TestA() = default;
};
class TestB final : public TestBase {
public:
TestB(int x, int y) : TestBase(x, y) {}
int getX() const override { return -TestBase::getX(); }
};
std::unique_ptr<ITest> constructTest(int x) {
// make_unique is c++14.
// For C++11 use std::unique_ptr<ITest>(new TestB(x, x)
if (x) {
return std::make_unique<TestB>(x, x);
// return std::unique_ptr<ITest>(new TestB(x, x));
}
return std::make_unique<TestA>();
}
The answer was to make the destructor a pure virtual AND to implement it with an empty body. That empty implementation is where I got tripped up. Print() doesn't need to be static.
Product.hpp
#include <memory>
class Product {
public:
virtual ~Product() = 0;
void Print();
protected:
char type_name{};
char size_name{};
private:
};
Product.cpp
#include "Product.hpp"
Product::~Product() {}
void Product::Print() {
//Print p
}
suppose I have those 2 classes:
class num{
public:
int a;
num(){};
num(int x):a(x){};
num(const num& n):a(n.a){}
virtual bool operator==(const num& n)const = 0;
virtual ~num(){};
};
class tmp: public num{
public:
tmp(){};
tmp(int x):num(x){};
tmp(const num& n): num(n){}
tmp(const tmp& t): num(t.a){}
virtual bool operator==(const num& n)const{
return tmp(n).a == a;
}
virtual ~tmp(){};
};
and something like this in main:
int main() {
num* x = &get(...);
return 0;
}
get return a reference of type tmp (in this case. in general it returns reference to a type that inherits from num)
what I want to do is create another num* y that will point to a copy of *x so that if I change *y I won't change *x. and I can't quite figure out how to do this since num is abstract so I can't create an object of this type to make a copy.
ok and another question.
if I overload << operator:
std::ostream& operator<<(std::ostream& os, const tmp& t){
return os<<t.a<<endl;
}
and then try to do this:
cout<<*x<<endl;
I get an error no match for 'operator<<'
why is that?
You're looking for the prototype pattern, also often called clone pattern.
It's basically a pure virtual method
virtual std::unique_ptr<num> clone() const = 0;
... that you declare in num, to be overriden by each derived class. You then just call x->clone(); to get a brand new object of the correct type.
You'll need a virtual function to clone an object based on its dynamic type. It will have to return a (preferably smart) pointer to a newly allocated object of the correct type. For example:
class num {
public:
virtual std::unique_ptr<num> clone() const = 0;
// other members...
};
class my_num : public num {
public:
virtual std::unique_ptr<num> clone() const {
return std::make_unique<my_num>(*this);
}
// other members...
};
int main() {
num* x = &get(...);
auto y = x->clone();
}
As far as I know, templated virtual functions aren't allowed/possible due to the undefined size of the vtable.
On the other hand, virtual functions inside a class template which don't use the template type seem to be allowed, right?
What about a virtual function that doesn't use the template type as parameter or return type but works on data of the template type? Would that be valid C++?
I have already done some testing and it seems to work.
My Code looks like this:
(Note: For reasons of readability this is only the basic structure, not the real code).
template<typename T>
class Base {
public:
virtual bool compare(void) {
// Basic implementation
return ((value1 + value2) < value3);
}
protected:
T value1, value2, value3;
}
/**
* Derived from Base<ComplexClass> where
* ComplexClass is a Class providing
* a int Value through .getInt()
**/
class Derived : Base<ComplexClass> {
bool compare(void) {
return ((value1.getInt() + value2.getInt()) < value3.getInt());
}
}
main {
Base<int> *intBase = new Base<int>();
Base<double> *doubleBase = new Base<double>();
Base<ComplexClass> *complexBase = new Derived();
intBase->compare(); // Should call base function
doubleBase->compare(); // Should also call base function
complexBase->compare(); // Should call the compare function of Derived
}
As far as i can tell this works like I excepted. Is this just a lucky coincidence or is this valid/good C++ style?
If it's valid, could someone please explain what's happening inside and why some people say it's forbidden/bad practice to derive from class templates and use virtual functions inside of class templates?
Thank you in advance!
PS: I know something similar could have been done by template specialization but I'd like to know if it's also possible this way.
Q As far as I know, templated virtual functions aren't allowed/possible due to the undefined size of the vtable.
A You can have virtual function in class templates.
Example code that compiles and links:
template <typename T>
struct Base
{
virtual T doSomething(T const& in) = 0;
Base(T const& data) : data_(data) {}
T data_;
};
struct Concrete : public Base<int>
{
Concrete(int d) : Base(d) {}
virtual int doSomething(int const& in)
{
return data_*in;
}
};
int main()
{
Concrete a(20);
int b = a.doSomething(10);
}
Q On the other hand, virtual functions inside a class template which don't use the template type seem to be allowed, right?
A The virtual functions of a class template can use anything -- not restricted to not using the template tye.
My example should make that clear.
Q What about a virtual function that doesn't use the template type as parameter or return type but works on data of the template type? Would that be valid C++?
A Yes, it will.
Again, my example should make that clear.
EDIT: Extended example
template <typename T>
struct Base
{
virtual T fun1(T const& in) = 0;
virtual T fun2(int in) = 0;
virtual int fun3(T const& in) = 0;
virtual int fun4(int in) = 0;
Base(T const& data) : data_(data) {}
T data_;
};
struct Concrete : public Base<int>
{
Concrete(int d) : Base(d) {}
virtual int fun1(int const& in)
{
return data_*in;
}
virtual int fun2(int in)
{
return fun1(in);
}
virtual int fun3(int const& in)
{
return fun1(in);
}
virtual int fun4(int in)
{
return fun1(in);
}
};
int main()
{
Concrete a(20);
int b = a.fun1(10);
int c = a.fun2(10);
int d = a.fun3(10);
int e = a.fun4(10);
}
This is perfectly valid. However, here you can have the same behaviour with specialization or just overloading, e.g.
template<typename T>
struct Base
{
bool compare() const { return val(value1) + val(value2) < val(value3); }
protected:
T value1, value2, value3;
private:
template<typename U>
static U val(U a) { return a; }
static int val(const ComplexClass& a) { return a.getInt(); }
};
Better keep virtual functions for when it's really needed.
And try to gather as much as possible shared code in a single place, minimizing what is to be specialized.
I have the following template classes,
I need to figure out how to implement a conversion operator between the derived template classes.
template<class T>
class Base
{
public:
Base () { }
template <class U>
operator Base<U>()
{
return Base<U> (v);
}
virtual double getV() = 0;
};
class D1: public Base<D1>
{
public:
D1(int j)
{
i = j;
}
double getV() const { return i; }
template <class U>
operator Base<U>()
{
return Base<U>(getV());
}
private:
int i;
};
class D2: public Base<D2>
{
public:
D2(int j)
{
i2 = j;
}
double getV() const { return i2; }
template <class U>
operator Base<U>()
{
return Base<U>(getV());
}
private:
int i2;
};
How can I achieve the following?
D1 d1(3);
D2 d2 = d1; //conversion from 'D1' to non-scalar type 'D2' requested
if the design itself sound or should I be doing something else?
Please let me know what ur thoughts
In your example, I don't see a reason why CRTP is used.
The specializations of Base all have a virtual member function that is not dependent on the template parameter. Your code suggests this virtual member function can be used to access all data necessary to create an instance of any class derived from a specialization of Base. If we follow this assumption, one could rather think of:
class Base
{
public:
virtual double getV() const = 0;
};
class D1 : public Base
{
int i;
public:
D1(int);
virtual double getV() const { return i; }
};
class D2 : public Base
{
int i;
public:
D2(int);
virtual double getV() const { return i; }
};
This still does not allow conversions. However, it is quite simple to add them here:
class D1 : public Base
{
int i;
public:
D1(int);
D1(Base const& p) : D1(p.getV()) {}
virtual double getV() const { return i; }
};
This conversion should be allowed by D1 and not by Base, because only D1 knows what data is required for it to be constructed.
If CRTP is necessary for things not shown here, you could still use a common base class:
class Common_base
{
public:
virtual double getV() const = 0;
};
template<class T>
class Base : public Common_base
{
};
class D1 : public Base<D1>
{
int i;
public:
D1(int);
D1(Common_base const& p) : D1(p.getV()) {}
virtual double getV() const { return i; }
};
If, for some reason, the CRTP is required for the conversion, you could still use a converting constructor template:
template<class T>
class Base
{
public:
virtual double getV() const = 0; // whyever
};
class D1 : public Base
{
int i;
public:
D1(int);
template<class U>
D1(Base<U> const& p) : D1(p.getV()) {}
virtual double getV() const { return i; }
};
It's hard to tell what you are trying to do but I don't think you can do it as described.
If you really want to have a common interface, you need to declare a common base class that includes all the methods and properties that your code needs, and have both classes derive from that. You can then always cast to the base class.
This seems ideal in fact. Especially since you can always use virtual methods to customize behavior in the base class if needed.
If that doesn't work within your current task, perhaps you should talk more about why you need this.
You either write a constructor for D2 that takes a const D1& or you write a conversion operator in D1 that returns a D2. You have to decide what it means to do this conversion and implement the conversion appropriately.
You can add a constructor in your derived classes that takes a Base template :
template <class U>
D2(const Base<U> &other)
{
i2 = other.getV();
}
But not sure if it will fit your needs.
I have implemented the following interface:
template <typename T>
class Variable
{
public:
Variable (T v) : m_value (v) {}
virtual void Callback () = 0;
private:
T m_value;
};
A proper derived class would be defined like this:
class Derived : public Variable<int>
{
public:
Derived (int v) : Variable<int> (v) {}
void Callback () {}
};
However, I would like to derive classes where Callback accepts different parameters (eg: void Callback (int a, int b)).
Is there a way to do it?
This is a problem I ran in a number of times.
This is impossible, and for good reasons, but there are ways to achieve essentially the same thing. Personally, I now use:
struct Base
{
virtual void execute() = 0;
virtual ~Base {}
};
class Derived: public Base
{
public:
Derived(int a, int b): mA(a), mB(b), mR(0) {}
int getResult() const { return mR; }
virtual void execute() { mR = mA + mB; }
private:
int mA, mB, mR;
};
In action:
int main(int argc, char* argv[])
{
std::unique_ptr<Base> derived(new Derived(1,2));
derived->execute();
return 0;
} // main
Even if such a thing were possible, it no longer makes much sense to have it as a virtual function, as the derived instantiations couldn't be called polymorphically via a pointer to the base class.
don't think this will be possible, because you can never interface it back to Variable.
This is what i mean
int a=0; int b = 0;
Variable<int>* derived = new Derived();
derived->Callback(a, b); //this won't compile because Variable<int> does not have Callback with 2 vars.
I know this there is an accepted answer, but there is one (ugly) way to achieve what you want, although I would not recommend it:
template <typename T>
class Variable
{
public:
Variable (T v) : m_value (v) {}
virtual void Callback (const char *values, ...) = 0;
private:
T m_value;
};
class Derived : public Variable<int>
{
public:
Derived (int v) : Variable<int> (v) {}
virtual void Callback (const char *values, ...) {
}
};
Now, you can use:
int a=0;
double b = 0;
Variable<int>* derived = new Derived(3);
derived->Callback("");
derived->Callback("df", a, b);
You need the values argument in order to obtain the remaining arguments inside the method. You also need to know the argument types, and pass them like printf does.
This method is error prone, as you must match the argument types on values with the real argument types.
You will have to add an overload of Callback in the base class that accepts these parameters. It would also be possible to do bad things, like accept a void*, or pass in a raw pointer-to-bytes.
The only scenario in which it is valid to alter virtual function signature is when you override the return value to something polymorphic to the original return value, e.g. *this.