I have an object S. S is composed of layers S0, S1, S2 ... just like a stack of stackable drawers.
I want to create a chain of template classes A, B, C such that:
They represents proxies to different layers of the S object.
Even during template instantiation, C can convert to B, which can convert to A.
A, B, and C have different sets of methods.
The problem is that if I use public inheritance, then C will get the methods of A and B.
Test:
#include <iostream>
// Library
template <typename T>
class A {
public:
void a() {std::cout << "a\n"; }
int s_{0};
};
template <typename T>
class B : public A<T> {
public:
void b() {std::cout << "b\n"; }
};
template <typename T>
class C : public B<T> {
public:
void c() {std::cout << "c\n"; }
};
// User of library write a function like this
template <typename T>
void foo(A<T>& a) { a.a(); }
// The problem:
int main() {
C<int> c;
foo(c);
c.a(); // <--- how to hide this?
return 0;
}
I'm not sure if I understand what you want. But one way to do it is changing the access level of base class member in derived class. For example:
template <typename T>
class C : public B<T> {
public:
void c() { std::cout << "c\n"; }
private:
using A::a; // <-- reduce access level of base class member
};
"Just" need 0 + 1 + 2 + ... + i conversion operators for layer Si
Or one template conversion operator if their attributes are all the same.
But this still need some way to control the conversions.
#include <iostream>
template <typename T>
class A {
public:
A(int& layer) : layer_(layer) {}
void a() {std::cout << "a\n"; }
int s_{0};
private:
int& layer_;
};
template <typename T>
class B {
public:
B(int& layer) : a_(layer) { }
template <template<typename> class X, typename T2>
operator X<T2>() { return a_; }
void b() {std::cout << "b\n"; }
private:
A<T> a_;
};
template <typename T>
class C {
public:
C(int& layer) : b_(layer) {}
template <template<typename> class X, typename T2>
operator X<T2>() { return b_; }
void c() {std::cout << "c\n"; }
private:
B<T> b_;
};
template <typename T>
class D {
public:
D(int& layer) : c_(layer) {}
template <template<typename> class X, typename T2>
operator X<T2>() { return c_; }
void c() {std::cout << "c\n"; }
private:
C<T> c_;
};
template <template<typename> class X, typename T>
void foo(X<T>& a) {
A<T>(a).a();
}
int main() {
int v = 1;
D<int> d(v);
foo(d);
return 0;
}
Related
class Base {};
class Derived : public Base {};
class SomeClass
{
template<typename T>
static void SetContent(T* pChild, OVariant content)
{
LOG_ASSERT(0, "All classes must be specialized!. Please provide implementation for this class.");
}
};
template <>
void SomeClass::SetContent(Base* value)
{
LOG_TRACE("Yay!");
}
int main() {
SomeClass foo;
Derived derived;
foo.SetContent(&derived);//want it to call SomeClass::SetContent(Base* value)
return 0;
}
When I call foo.SetContent(derived), I get the Assert. Is it not possible for the derived class to use the specialization for it's base class?
You can convert a Derived* to a Base*, but I think you rather want to specialize for all T that have Base as base
#include <type_traits>
#include <iostream>
class Base {};
class Derived : public Base {};
template <typename T,typename = void>
struct impl {
void operator()(T*) {
std::cout <<"All classes must be specialized!. Please provide implementation for this class.\n";
}
};
template <typename T>
struct impl<T,std::enable_if_t<std::is_base_of_v<Base,T>>> {
void operator()(T*) {
std::cout << "Yay\n";
}
};
class SomeClass
{
public:
template<typename T>
static void SetContent(T* pChild)
{
impl<T>{}(pChild);
}
};
struct bar{};
int main() {
SomeClass foo;
Derived derived;
foo.SetContent(&derived);
bar b;
foo.SetContent(&b);
}
Output:
Yay
All classes must be specialized!. Please provide implementation for this class.
//want it to call SomeClass::SetContent(Base* value)
Note that if the template argument is deduced, it will be deduced as Derived not as Base and the argument is Derived*. SomeClass::SetContent<Base>(&derived); would already work as expected with your code (because Derived* can be converted to Base*).
A workaround would be to have all SetContent's explicit specializations to form an overload set. You will have to do it yourself:
#include <iostream>
#include <utility>
#include <functional>
class Base {};
class Derived : public Base {};
template <class... T>
struct Overloads : T... {
Overloads(const T &... t) : T(t)... {}
using T::operator()...;
};
template <class R, class... Args>
struct FunctionP {
using F = R(Args...);
FunctionP(F *t) : t_(t) {}
R operator()(Args... args) const {
return std::invoke(t_, std::forward<Args>(args)...);
}
F *t_;
};
struct SomeClass {
template<typename T>
static void SetContent(T *x) {
Overloads o(FunctionP(&SetContentImpl<Base>)); // enumerates all the specializations here
if constexpr (std::is_invocable_v<decltype(o), T *>) {
o(x);
} else {
SetContentImpl(x);
}
}
template<typename T>
static void SetContentImpl(T *) {
std::cout << "1";
}
};
template <>
void SomeClass::SetContentImpl(Base *) {
std::cout << "2";
}
int main() {
SomeClass foo;
Derived derived;
foo.SetContent(&derived);//want it to call SomeClass::SetContent(Base* value)
return 0;
}
godbolt
I instance a template and want to use that instance for other template
template <typename I> class A
{
I name;
public:
A(I n){
name = n;
}
void show() const
{
cout << name << "\t";
}
};
template <template<typename I> class V> class B {
V origen;
public:
B (V o){
origen = o;
}
};
auto main() -> int
{
cout << "[code.cpp]" << endl;
A<int> a1(1);
a1.show();
B<A> b1(a1);
return 0;
}
What I see about this topic is that´s not usually use, but when use, not used in this form, and I don't understand why.
I tried using const. Not working
V origen; and void B (V o) are invalid as V is not a type.
Maybe you want
template <typename T> class B;
template <template<typename> class V, typename T>
class B<V<T>>
{
V<T> origen;
public:
B (V<T> o) : origen(o) {}
};
and then
A<int> a1(1);
B<A<int>> b1(a1);
Or maybe
template <template<typename> class V>
class B
{
V<int> origen;
public:
B (V<int> o) : origen(o) {}
};
and then
A<int> a1(1);
B<A> b1(a1);
I have a diamond inheritance scheme, where the last child should be able to inherit from many different parents.
A
/|\
/ | \
B C ...
| | |
* *
D E
Now imagine I have a class D : public B, class E : public B, public C, etc. From D I want to call the same function of all its parents, which I am guaranteed exists due to the inheritance. My thought was that I could wrap this in some variadic template.
Currently I have this:
template <typename T>
class A
{
public:
A(T t) : mT(t) {}
virtual ~A() {}
virtual void doThings() = 0;
protected:
T mT;
};
template <typename T, typename A = A<T>>
class B : public A
{
public:
B(T t) : A(t) {}
virtual ~B() {}
virtual void doThings() { std::cout << "B" << std::endl; }
};
template <typename T, typename A = A<T>>
class C : public A
{
public:
C(T t) : A(t) {}
virtual ~C() {}
virtual void doThings() { std::cout << "C" << std::endl; }
};
Now I thought I could do something like this, which obviously does not work:
template <typename T, typename ...Args>
class ChildGenerator : public Args...
{
public:
ChildGenerator(T t) : Args(t)... {}
// The unpacking of the variadic template does not work here.
// Do I need to make it recursive somehow? How can I do that without having to instantiate new classes B and C?
void doThings() override { Args...::doThings();}
};
My hope is that I can use it like so:
int main()
{
using B = B<double>;
using C = C<double>;
B c1(0.0);
C c2(1.0);
ChildGenerator<double, B, C> c3(2.0);
c1.doThings();
c2.doThings();
c3.doThings();
}
Expected output (order does not matter):
B
C
B // <-- order of these two does not matter
C // <--
Is what I'm trying to achieve possible?
One way to iterate over the variadic bases:
template <typename T, typename ...Args>
class ChildGenerator : public Args...
{
public:
ChildGenerator(T t) : Args(t)... {}
void doThings() override {
int dummy[] = {0, (Args::doThings(), void(), 0)...};
static_cast<void>(dummy); // avoid warning for unused variable
}
};
or in C++17, with folding expression:
void doThings() override {
(static_cast<void>(Args::doThings()), ...);
}
Use a fold-expression (C++17):
void doThings() override { ((Args::doThings()) , ...);}
Live demo
I have a diamond inheritance scheme, where the last child should be able to inherit from many different parents.
A
/|\
/ | \
B C ...
| | |
* *
D E
Now imagine I have a class D : public B, class E : public B, public C, etc. From D I want to call the same function of all its parents, which I am guaranteed exists due to the inheritance. My thought was that I could wrap this in some variadic template.
Currently I have this:
template <typename T>
class A
{
public:
A(T t) : mT(t) {}
virtual ~A() {}
virtual void doThings() = 0;
protected:
T mT;
};
template <typename T, typename A = A<T>>
class B : public A
{
public:
B(T t) : A(t) {}
virtual ~B() {}
virtual void doThings() { std::cout << "B" << std::endl; }
};
template <typename T, typename A = A<T>>
class C : public A
{
public:
C(T t) : A(t) {}
virtual ~C() {}
virtual void doThings() { std::cout << "C" << std::endl; }
};
Now I thought I could do something like this, which obviously does not work:
template <typename T, typename ...Args>
class ChildGenerator : public Args...
{
public:
ChildGenerator(T t) : Args(t)... {}
// The unpacking of the variadic template does not work here.
// Do I need to make it recursive somehow? How can I do that without having to instantiate new classes B and C?
void doThings() override { Args...::doThings();}
};
My hope is that I can use it like so:
int main()
{
using B = B<double>;
using C = C<double>;
B c1(0.0);
C c2(1.0);
ChildGenerator<double, B, C> c3(2.0);
c1.doThings();
c2.doThings();
c3.doThings();
}
Expected output (order does not matter):
B
C
B // <-- order of these two does not matter
C // <--
Is what I'm trying to achieve possible?
One way to iterate over the variadic bases:
template <typename T, typename ...Args>
class ChildGenerator : public Args...
{
public:
ChildGenerator(T t) : Args(t)... {}
void doThings() override {
int dummy[] = {0, (Args::doThings(), void(), 0)...};
static_cast<void>(dummy); // avoid warning for unused variable
}
};
or in C++17, with folding expression:
void doThings() override {
(static_cast<void>(Args::doThings()), ...);
}
Use a fold-expression (C++17):
void doThings() override { ((Args::doThings()) , ...);}
Live demo
I wonder whether the below tricky situation is possible:
Suppose I have a template class template <typename DTYPE> class A{};, where DTYPE is supposed to be one of uint8_t, uint16_t, etc. I want to add a friend class to A, but this friend class differs for each DTYPE alternative. Further, suppose the friend classes for different DTYPE values are not instantiations of another template class, but independent classes.
Is there a way to do it?
You can add template "proxy" class FriendOfA and specialize it for whatever type you need:
// actual friends
class FriendUint8 {};
class FriendUint16 {};
template<typename T> struct FriendOfA;
template<>
struct FriendOfA<uint8_t> {
typedef FriendUint8 type;
};
template<>
struct FriendOfA<uint16_t> {
typedef FriendUint16 type;
};
// optional helper
template <typename T>
using FriendOfA_t = typename FriendOfA<T>::type;
template<class T>
class A {
friend typename FriendOfA<T>::type;
// or simply
friend FriendOfA_t<T>;
};
I believe you're looking for something like that:
#include <iostream>
struct BaseFriend
{
template <typename T>
void boo(const T& t) { t.foo(); }
};
struct BaseFriendProxy
{
template <typename T>
void boo(const T& t) { std::cout << "Proxy: "; t.foo(); }
};
template <typename TType>
struct Friend ;
template <>
struct Friend<int> {
using T = BaseFriend;
};
template <>
struct Friend<char> {
using T = BaseFriendProxy;
};
template <typename DTYPE>
class A
{
private:
friend typename Friend<DTYPE>::T;
void foo() const
{ std::cout << "A::foo()" << std::endl; }
};
int main()
{
A<int> a;
BaseFriend bf1;
bf1.boo(a);
A<char> b;
BaseFriendProxy bf2;
bf2.boo(b);
return 0;
}
But this will work only with C++11: you can't combine friend class ... with typename X::Y in C++03
Sure you can, specialize your template and add whatever friend class you want:
#include <iostream>
using namespace std;
class base {
public:
virtual int getValue() = 0;
};
class friend1 {
public:
base* ptr;
int getValue() {
return ptr->getValue();
}
};
class friend2 {
public:
base* ptr;
int getValue() {
return ptr->getValue();
}
};
template <typename DTYPE> class A : public base{
public:
A() { data = 42; }
// No friends
private:
int data;
int getValue() {
return data;
}
};
template <> class A<char> : public base{
public:
A() { data = 44; }
friend class friend1;
private:
int data;
int getValue() {
return data;
}
};
template <> class A<bool> : public base{
public:
A() { data = 45; }
friend class friend2;
private:
int data;
int getValue() {
return data;
}
};
int main()
{
A<char> obj1;
friend1 friend_of_obj1;
friend_of_obj1.ptr = &obj1;
cout << friend_of_obj1.getValue() << endl;
A<bool> obj2;
friend2 friend_of_obj2;
friend_of_obj2.ptr = &obj2;
cout << friend_of_obj2.getValue();
}
http://ideone.com/hM9x0y
Yes, a friend can be based on a template. Such as;
template <typename DTYPE>
struct Friend;
template <class DTYPE>
class A {
friend struct Friend<DTYPE>;
};
For each type DTYPE for the class A you wish to support (different by implementation) you can specialise Friend, such as;
template<>
struct Friend<uint8_t> {
// ...
};
A basic working sample:
#include <cstdint>
using namespace std;
template<typename T>
struct Friend;
template <class T>
class A {
friend struct Friend<T>;
int i = 0;
};
template<>
struct Friend<uint8_t> {
void method() {
A<uint8_t> a;
a.i = 8;
}
};
template<>
struct Friend<uint16_t> {
void method() {
A<uint16_t> a;
//A<uint8_t> b; // fails to compile
a.i = 16;
}
};
int main()
{
A<uint8_t> a;
Friend<uint8_t> f;
f.method();
}