shared_ptr 的隐式转换 [英] Implicit conversion of shared_ptr

问题描述

我有两个类 U 和 T 的 shared_ptr，其中 T 是 U 的基础.

I have two shared_ptrs of the classes U and T where T is the base of U.

从shared_ptr到shared_ptr的隐式转换是没有问题的.但是也可以将 shared_ptr 转换为 shared_ptr ?

It is no problem to do an implicit conversion from shared_ptr<U> to shared_ptr<T>. But is is also possible to do a conversion from shared_ptr<T> to shared_ptr<U> ?

我尝试了建议的解决方案:

I tried the sugested solution:

class T {
public:
  virtual ~T() {}
protected: 
  void fillData() = 0;
};

class Timpl : public T 
{ 
public:
  virtual ~Timpl() {}
protected:
  virtual void fillData() = 0;
};

class U : public Timpl {
public: 
  virtual ~U() {}
protected: 
  virtual void fillData() {...} // fill 
};

typedef  shared_ptr<T> TPtr
typedef  shared_ptr<U> UPtr


TPtr tp = std::make_shared<U>();  
UPtr up = std::static_pointer_cast<U>(tp);    //<-- error here :

错误:没有匹配的函数调用'static_pointer_cast(TPtr)'

error: no matching function for call to 'static_pointer_cast(TPtr)'

注意:模板 std::__shared_ptr<_Tp1, _Lp> std::static_pointer_cast(const std::__shared_ptr<_Tp2, _Lp>&)

note: template std::__shared_ptr<_Tp1, _Lp> std::static_pointer_cast(const std::__shared_ptr<_Tp2, _Lp>&)

注意:模板参数推导/替换失败:

note: template argument deduction/substitution failed:

注意:'TPtr {aka boost::shared_ptr}' 不是从 'const std::__shared_ptr<_Tp2, _Lp>' 派生的

note: 'TPtr {aka boost::shared_ptr}' is not derived from 'const std::__shared_ptr<_Tp2, _Lp>'

此问题的解决方案:

将 std::shared_ptr 与 boost::shared_ptr 混合起来不是个好主意.

mixing of std::shared_ptr with boost::shared_ptr is no good idea.

推荐答案

是的，使用static_pointer_cast:

#include <memory>

struct T { virtual ~T() {} };
struct U : T {};

std::shared_ptr<T> pt = std::make_shared<U>();     // *pt is really a "U"

auto pu = std::static_pointer_cast<U>(pt);

还有一个匹配的 std::dynamic_pointer_cast 如果无法转换，它会返回一个空指针.

There's also a matching std::dynamic_pointer_cast which returns a null pointer if the conversion isn't possible.

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