实现智能指针类时出错

Question

我被一个错误困住了。这是我的代码：

 template<class t>
 class smart_ptr{
    t *ptr;
 public: 
    smart_ptr(t *p):ptr(p){cout<<"smart pointer copy constructor is called"<<endl;} 

    ~smart_ptr(){cout<<"smart pointer destructor is called"<<endl;delete(ptr);}
    t& operator *(){cout<<"returning the * of pointer"<<endl;return(*ptr);}
    t* operator ->(){cout<<"returning the -> of pointer"<<endl;return(ptr);}
    t* operator=(const t &lhs){ptr=lhs;cout<<"assignement operator called"<<endl;}
   };
   class xxx{
            int x;
    public:
            xxx(int y=0):x(y){cout<<"xxx constructor called"<<endl;}
            ~xxx(){cout<<"xxx destructor is called"<<endl;}
            void show(){cout<<"the value of x="<<x<<endl;}
    };
 int main(int argc, char *argv[])
 {
    xxx *x1=new xxx(50);
    smart_ptr<xxx *> p1(x1);
    return 0;
 }

在编译过程中，我会遇到以下错误

smart_pointer_impl.cpp:在函数‘int(int，char**)’中：

smart_pointer_impl.cpp:27:错误:没有调用‘smart_ptr：：smart_ptr(xxx*&)’的匹配函数

smart_pointer_impl.cpp:7:注:候选人是: smart_ptr::smart_ptr(t*)，t= xxx*

smart_pointer_impl.cpp:4:注: smart_ptr::smart_ptr(const smart_ptr&)

任何解决方案的帮助都是最受欢迎的。

Answer 1

你需要改变：

smart_ptr<xxx *> p1(x1); to smart_ptr<xxx> p1(x1); 

看起来不错。

Answer 2

据推测，smart_ptr是template<class t>，而总体上是smart_ptr<xxx>而不是smart_ptr<xxx*>

Answer 3

您还没有将类smart_ptr声明为模板

template <TYPE>
class smart_ptr{
    TYPE *ptr;
public: 
    smart_ptr(TYPE *p):ptr(p){cout<<"smart pointer copy constructor is called"<<endl;} 

    ~smart_ptr(){cout<<"smart pointer destructor is called"<<endl;delete(ptr);}
    TYPE& operator *(){cout<<"returning the * of pointer"<<endl;return(*ptr);}
    TYPE* operator ->(){cout<<"returning the -> of pointer"<<endl;return(ptr);}
    TYPE* operator=(const TYPE &lhs){ptr=lhs;cout<<"assignement operator called"<<endl;}
};

另外，您正在声明您的指针类型为‘指针到xxx’，但是模板类是指向类型的指针。尝试：

smart_ptr<xxx> p1(x1);