如何设置表示接口的类?这只是一个抽象基类吗?


当前回答

如果您只需要接口的静态绑定(没有虚拟的,没有接口类型本身的实例,接口仅作为指南):

#include <iostream>
#include <string>

// Static binding interface
// Notice: instantiation of this interface should be usefuless and forbidden.
class IBase {
 protected:
  IBase() = default;
  ~IBase() = default;

 public:
  // Methods that must be implemented by the derived class
  void behaviorA();
  void behaviorB();

  void behaviorC() {
    std::cout << "This is an interface default implementation of bC().\n";
  };
};

class CCom : public IBase {
  std::string name_;

 public:
  void behaviorA() { std::cout << "CCom bA called.\n"; };
};

class CDept : public IBase {
  int ele_;

 public:
  void behaviorB() { std::cout << "CDept bB called.\n"; };
  void behaviorC() {
    // Overwrite the interface default implementation
    std::cout << "CDept bC called.\n";
    IBase::behaviorC();
  };
};

int main(void) {
  // Forbid the instantiation of the interface type itself.
  // GCC error: ‘constexpr IBase::IBase()’ is protected within this context
  // IBase o;

  CCom acom;
  // If you want to use these interface methods, you need to implement them in
  // your derived class. This is controled by the interface definition.
  acom.behaviorA();
  // ld: undefined reference to `IBase::behaviorB()'
  // acom.behaviorB();
  acom.behaviorC();

  CDept adept;
  // adept.behaviorA();
  adept.behaviorB();
  adept.behaviorC();
  // adept.IBase::behaviorC();
}

其他回答

如果您只需要接口的静态绑定(没有虚拟的,没有接口类型本身的实例,接口仅作为指南):

#include <iostream>
#include <string>

// Static binding interface
// Notice: instantiation of this interface should be usefuless and forbidden.
class IBase {
 protected:
  IBase() = default;
  ~IBase() = default;

 public:
  // Methods that must be implemented by the derived class
  void behaviorA();
  void behaviorB();

  void behaviorC() {
    std::cout << "This is an interface default implementation of bC().\n";
  };
};

class CCom : public IBase {
  std::string name_;

 public:
  void behaviorA() { std::cout << "CCom bA called.\n"; };
};

class CDept : public IBase {
  int ele_;

 public:
  void behaviorB() { std::cout << "CDept bB called.\n"; };
  void behaviorC() {
    // Overwrite the interface default implementation
    std::cout << "CDept bC called.\n";
    IBase::behaviorC();
  };
};

int main(void) {
  // Forbid the instantiation of the interface type itself.
  // GCC error: ‘constexpr IBase::IBase()’ is protected within this context
  // IBase o;

  CCom acom;
  // If you want to use these interface methods, you need to implement them in
  // your derived class. This is controled by the interface definition.
  acom.behaviorA();
  // ld: undefined reference to `IBase::behaviorB()'
  // acom.behaviorB();
  acom.behaviorC();

  CDept adept;
  // adept.behaviorA();
  adept.behaviorB();
  adept.behaviorC();
  // adept.IBase::behaviorC();
}

除了上面写的内容,还有一点补充:

首先,确保析构函数也是纯虚拟的

第二,您可能希望在执行时实际上继承(而不是正常继承),只是为了获得好的度量。

在C++20中,可以使用概念而不是类。它比继承更有效率。

template <class T>
concept MyInterface = requires (T t) {
    { t.interfaceMethod() };
};

class Implementation {
public:
    void interfaceMethod();
};
static_assert(MyInterface<Implementation>);

然后您可以在函数中使用它:

void myFunction(MyInterface auto& arg);

限制是不能在容器中使用它。

就我所能测试的而言,添加虚拟析构函数非常重要。我使用的是用new创建的对象,用delete销毁的对象。

如果不在接口中添加虚拟析构函数,则不会调用继承类的析构函数。

class IBase {
public:
    virtual ~IBase() {}; // destructor, use it to call destructor of the inherit classes
    virtual void Describe() = 0; // pure virtual method
};

class Tester : public IBase {
public:
    Tester(std::string name);
    virtual ~Tester();
    virtual void Describe();
private:
    std::string privatename;
};

Tester::Tester(std::string name) {
    std::cout << "Tester constructor" << std::endl;
    this->privatename = name;
}

Tester::~Tester() {
    std::cout << "Tester destructor" << std::endl;
}

void Tester::Describe() {
    std::cout << "I'm Tester [" << this->privatename << "]" << std::endl;
}


void descriptor(IBase * obj) {
    obj->Describe();
}

int main(int argc, char** argv) {

    std::cout << std::endl << "Tester Testing..." << std::endl;
    Tester * obj1 = new Tester("Declared with Tester");
    descriptor(obj1);
    delete obj1;

    std::cout << std::endl << "IBase Testing..." << std::endl;
    IBase * obj2 = new Tester("Declared with IBase");
    descriptor(obj2);
    delete obj2;

    // this is a bad usage of the object since it is created with "new" but there are no "delete"
    std::cout << std::endl << "Tester not defined..." << std::endl;
    descriptor(new Tester("Not defined"));


    return 0;
}

如果在没有virtual~IBase(){};的情况下运行前面的代码;,您将看到从未调用析构函数Tester::~Tester()。

我还是C++开发的新手。我从Visual Studio(VS)开始。

然而,似乎没有人提到VS(.NET)中的__interface。我不太确定这是否是声明接口的好方法。但它似乎提供了额外的强制执行(文件中提到)。这样就不必显式指定虚拟TYPE Method()=0;,因为它将被自动转换。

__interface IMyInterface {
   HRESULT CommitX();
   HRESULT get_X(BSTR* pbstrName);
};

然而,我不使用它,因为我担心跨平台编译兼容性,因为它只在.NET下可用。

如果有人对此感兴趣,请分享。:-)

谢谢