给定
class A {
public:
virtual int foo (int) const = 0;
virtual void bar (char, double) const = 0;
};
class B : public A {
virtual int foo (int) const {std::cout << "B::foo() called.\n"; return 3;}
virtual void bar () const {std::cout << "B::bar() called.\n";}
};
class C : public B {
virtual int foo (int) const {std::cout << "C::foo() called.\n"; return 8;}
virtual void bar (char, double) const {std::cout << "C::bar() called.\n";}
};
我想将
foo
和bar
(以及A
的其他虚拟函数)放入函数的模板系列中。到目前为止,这是我想到的:#include <iostream>
enum Enum {Foo, Bar};
template <Enum> struct EnumTraits;
template <> struct EnumTraits<Foo> { using return_type = int; };
template <> struct EnumTraits<Bar> { using return_type = void; };
class A {
template <Enum> class Execute;
public:
virtual int foo (int) const = 0;
virtual void bar (char, double) const = 0;
template <Enum E, typename... Args>
typename EnumTraits<E>::return_type execute(Args&&... args) const {
return Execute<E>(this)(std::forward<Args>(args)...);
}
};
template <>
class A::Execute<Foo> {
const A* a;
public:
Execute (const A* a_) : a(a_) {}
template <typename... Args>
int operator()(Args&&... args) const {return a->foo(std::forward<Args>(args)...);}
};
template <>
class A::Execute<Bar> {
const A* a;
public:
Execute (const A* a_) : a(a_) {}
template <typename... Args>
void operator()(Args&&... args) const {a->bar(std::forward<Args>(args)...);}
};
class B : public A {
virtual int foo (int) const {std::cout << "B::foo() called.\n"; return 3;}
virtual void bar () const {std::cout << "B::bar() called.\n";}
};
class C : public B {
virtual int foo (int) const {std::cout << "C::foo() called.\n"; return 8;}
virtual void bar (char, double) const {std::cout << "C::bar() called.\n";}
};
int main() {
A* c = new C;
int n = c->foo(5); // C::foo() called.
c->bar(3, 'c'); // C::bar() called.
n = c->execute<Foo>(5); // C::foo() called.
c->execute<Bar>(3, 'c'); // C::bar() called.
}
但是,特化
A::Execute<Foo>
和A::Execute<Bar>
看起来几乎相同,并且理想情况下应保留非特化(特别是如果存在foo
和bar
之外的许多其他虚函数)。写类似:template <Enum N>
class A::Execute {
const A* a;
public:
Execute (const A* a_) : a(a_) {}
template <typename... Args>
int operator()(Args&&... args) const {return a->???(std::forward<Args>(args)...);}
};
如何填写???部分?理想情况下,我希望使用已经存在的
EnumTraits
类。 最佳答案
这是我的尝试。我已将enum
替换为用作标签的struct
,并将EnumTraits
替换为TagTraits
。我喜欢struct
方法,因为它允许添加新标签而不影响现有标签。
#include <iostream>
#include <functional>
template <typename T> struct TagTraits;
// Generic implementation of A based on TagTraits.
class A {
template <typename Tag, typename... Args>
class Execute {
const A* a;
public:
Execute (const A* a_) : a(a_) {}
typename TagTraits<Tag>::return_type operator()(Args&&... args) const
{
return (a->*(TagTraits<Tag>::get_funtion_ptr()))(std::forward<Args>(args)...);
}
};
public:
virtual int foo (int) const = 0;
virtual void bar (char, double) const = 0;
template <typename Tag, typename... Args>
typename TagTraits<Tag>::return_type execute(Args&&... args) const
{
return Execute<Tag, Args...>(this)(std::forward<Args>(args)...);
}
};
// tag for foo and the corresponding TagTraits
struct foo_tag {};
template <> struct TagTraits<foo_tag>
{
using return_type = int;
static decltype(&A::foo) get_funtion_ptr(){ return &A::foo;}
};
// tag for bar and the corresponding TagTraits
struct bar_tag {};
template <> struct TagTraits<bar_tag>
{
using return_type = void;
static decltype(&A::bar) get_funtion_ptr(){ return &A::bar;}
};
// Derived classes of A.
class B : public A {
virtual int foo (int) const {std::cout << "B::foo() called.\n"; return 3;}
virtual void bar (char, double) const {std::cout << "B::bar() called.\n";}
};
class C : public B {
virtual int foo (int) const {std::cout << "C::foo() called.\n"; return 8;}
virtual void bar (char, double) const {std::cout << "C::bar() called.\n";}
};
// Test B
void test_B()
{
A* aPtr = new B;
int n = aPtr->foo(5); // B::foo() called.
aPtr->bar(3, 'c'); // B::bar() called.
n = aPtr->execute<foo_tag>(5); // B::foo() called.
aPtr->execute<bar_tag>(3, 'c'); // B::bar() called.
}
// Test C
void test_C()
{
A* aPtr = new C;
int n = aPtr->foo(5); // C::foo() called.
aPtr->bar(3, 'c'); // C::bar() called.
n = aPtr->execute<foo_tag>(5); // C::foo() called.
aPtr->execute<bar_tag>(3, 'c'); // C::bar() called.
}
int main()
{
test_B();
test_C();
}
输出:
B::foo() called.
B::bar() called.
B::foo() called.
B::bar() called.
C::foo() called.
C::bar() called.
C::foo() called.
C::bar() called.
关于c++ - 将虚拟功能融入家庭,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/32240503/