C++ unique_ptr <Base>指向Derived的ptr

Question

我有以下代码:

class Base{
public:
    virtual void do_something(int a) = 0;
    virtual ~Base();

};

template<typename T>
class Concrete : public Base{

    T _t;
public:
    Concrete(){};
    virtual void do_something(int a);
    virtual ~Concrete();
};

template<typename T>
void Concrete<T>::do_something(int a){
    std::cout << a << std::endl;
}
template<typename T>
Concrete<T>::~Concrete(){}

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


    std::unique_ptr<Base> b(new Concrete<int>());
    b->do_something(5);
}

但是,链接退出时出错.输出是:

Undefined symbols for architecture x86_64:
  "Base::~Base()", referenced from:
      Concrete<int>::~Concrete() in testbed_evaluator.o
      Concrete<int>::~Concrete() in testbed_evaluator.o
  "typeinfo for Base", referenced from:
      typeinfo for Concrete<int> in testbed_evaluator.o
ld: symbol(s) not found for architecture x86_64
collect2: error: ld returned 1 exit status

我无法弄清楚我做错了什么

Answer 1

你向编译器承诺了一个析构函数Base,所以它调用了它.但是你从来没有写过一个,所以当链接器看到那些调用它不知道该怎么做.

您可以像更改一样轻松地修复它

virtual ~Base();

至

virtual ~Base() {}

后者有一个身体; 这是一个真正的定义.