重载c++模板类方法

Question

我可以在扩展其特化的类中重载模板类函数吗？

我有以下一段代码（我已尝试将其简化到最低限度）：

#include <iostream>

using namespace std;

class X {
 public:
  unsigned test_x() {
    return 1;
  }
};

class Y {
 public:
  unsigned test_y() {
    return 2;
  }
};

template <typename T, typename U>
class A {

 public: 

  unsigned foo(U i) {
    cout << "A" << endl;   
    return i.test_x();
  }

  unsigned bar(T i) {
    return foo(i);
  }

};

class B : public A<Y, X> {
 public:
  unsigned foo(Y i) {
    cout << "B" << endl;   
    return i.test_y();
  }
};

int  main() {

  B b = B();
  Y y = Y();
  cout << "Hello: " << b.bar(y) << endl;   
  return 0;

}

但是编译器会产生以下错误：

hello.cc: In member function ‘unsigned int A<T, U>::bar(T) [with T = Y, U = X]’:
hello.cc:47:   instantiated from here
hello.cc:30: error: no matching function for call to ‘A<Y, X>::foo(Y&)’
hello.cc:24: note: candidates are: unsigned int A<T, U>::foo(U) [with T = Y, U = X]

基本上我想在其派生类 B 中重载函数 A::foo()。

Answer 1

显然你要问的是所谓的“静态多态性”，它是通过“奇怪地重复出现的模板模式”来实现的：

template <typename Derived, typename T, typename U>
class A {
 public:

  unsigned foo(U i) {
    cout << "A" << endl;   
    return i.test_x();
  }

  unsigned bar(T i) {
    return static_cast<Derived*>(this)->foo(i);
  }

};

class B : public A<B, Y, X> {
 public:
  // Uncomment this line if you really want to overload foo
  // instead of overriding. It's optional in this specific case.
  //using A::foo; 

  unsigned foo(Y i) {
    cout << "B" << endl;   
    return i.test_y();
  }
};

Answer 2

我不认为派生类中的重载不能从要调用它们的基类中使用。

你可以做的是提取foo方法，这样你就可以将它专门用于B.

#include <iostream>

using namespace std;

class X {
 public:
  unsigned test_x() {
    return 1;
  }
};

class Y {
 public:
  unsigned test_y() {
    return 2;
  }
};

template <typename T, typename U>
struct Foo
{
    unsigned foo(U i) {
    cout << "A" << endl;
    return i.test_x();
  }
};

template <typename T, typename U>
class A : public Foo<T, U> {

 public:
  unsigned bar(T i) {
    return this->foo(i);
  }
};

template <>
struct Foo<Y, X>
{
 public:
  unsigned foo(Y i) {
    cout << "B" << endl;
    return i.test_y();
  }
};

class B : public A<Y, X> {
};

int  main() {

  B b = B();
  Y y = Y();
  cout << "Hello: " << b.bar(y) << endl;
  return 0;
}

Answer 3

编辑：我的第一个答案不正确。

当您使用类Y 和X 实例化A 时，foo(T) 的调用会产生错误，因为A 中没有定义适当的重载方法。在A中声明一个foo(T)的纯虚方法并在B中实现这个方法就足够了。

template <typename T, typename U>
class A {
public: 
  virtual unsigned foo(T i) = 0;
  unsigned foo(U i) { /* as above */ }
  unsigned bar(T i) {
    return foo(i); // calls abstract foo(T)
  }
};
/* B is left untouched */

编译它会生成这个输出：

B
Hello: 2