c++ - 如何将 Base* 指针视为 Derived<T>* 指针？

Question

我想将指向 Base 类的指针存储在 vector 中，然后将它们用作函数参数，它们充当特定类，请参见此处:

#include <iostream>
#include <vector>

class Base {};

template<class T>
class Derived : public Base {};

void Foo(Derived<int>* d) {
  std::cerr << "Processing int" << std::endl;
}

void Foo(Derived<double>* d) {
  std::cerr << "Processing double" << std::endl;
}

int main() {
  std::vector<Base*> vec;
  vec.push_back(new Derived<int>());
  vec.push_back(new Derived<double>());
  Foo(vec[0]);
  Foo(vec[1]);
  delete vec[0];
  delete vec[1];
  return 0;
}

这不编译:

error: call of overloaded 'Foo(Base*&)' is ambiguous

有可能让它发挥作用吗？我需要根据 vector 的 int、double 等类型以不同方式处理 vector 的元素。

Answer 1

您需要使用方法多态性，因为它是动态的，而不是函数重载，后者是静态的(编译时)。要重载模板化类型，您需要使用模板特化。

例子:

#include <iostream>
#include <vector>

class Base {
  public:
    virtual void Foo() {
      std::cerr << "Processing base" << std::endl;
    }
};

template<class T>
class Derived : public Base {};

template <>
class Derived <int> : public Base {
  public:
    void Foo() {
      std::cerr << "Processing int" << std::endl;
    }
};

template <>
class Derived <double> : public Base {
  public:
    void Foo() {
      std::cerr << "Processing double" << std::endl;
    }
};

int main() {
  std::vector<Base*> vec;
  vec.push_back(new Derived<int>());
  vec.push_back(new Derived<double>());
  vec[0]->Foo();
  vec[1]->Foo();
  delete vec[0];
  delete vec[1];
  return 0;
}