重新设置shared_ptr<派生>;来自接受shared_ptr的函数

发布于 2025-01-14 00:41:52 字数 1103 浏览 3 评论 0原文

我正在尝试从接受基类的 shared_ptr 的函数重新设置派生类的 shared_ptr 。

这个答案是相关的，但它没有涵盖我需要重新安装指针以便不能使用常量引用的事实。

这是一个 MWE:

#include <memory>

class A {
};

class B : public A {
};

class C {
 public:
 void doSomethingWithA(std::shared_ptr<A>& a){
     a = std::make_shared<B>();
 }
    
};

int main()
{
  std::shared_ptr<B> b;
  C c;
  c.doSomethingWithA(b);
}

它给出了以下编译错误：

 In function 'int main()':
21:23: error: no matching function for call to 'C::doSomethingWithA(std::shared_ptr<B>&)'
21:23: note: candidate is:
11:7: note: void C::doSomethingWithA(std::shared_ptr<A>&)
11:7: note:   no known conversion for argument 1 from 'std::shared_ptr<B>' to 'std::shared_ptr<A>&'

代码背后的原因是可能有许多派生类，而我只在运行时知道要实例化哪一个。类C 将根据运行时发生的情况决定分配多种派生类型之一。

我不想创建 doSomethingWithA() 的多个重载，一般来说，将派生类的 shared_ptr 分配给 shared_ptr 是合法的> 基类的，所以我不明白为什么不能进行转换。

原文

I'm trying to reseat a shared_ptr of a derived class from a function that accept a shared_ptr of a base class.

This answer is relevant, but it does not cover the fact that I need to reseat the pointer so a const reference cannot be used.

Here is a MWE:

#include <memory>

class A {
};

class B : public A {
};

class C {
 public:
 void doSomethingWithA(std::shared_ptr<A>& a){
     a = std::make_shared<B>();
 }
    
};

int main()
{
  std::shared_ptr<B> b;
  C c;
  c.doSomethingWithA(b);
}

which gives the following compilation error:

 In function 'int main()':
21:23: error: no matching function for call to 'C::doSomethingWithA(std::shared_ptr<B>&)'
21:23: note: candidate is:
11:7: note: void C::doSomethingWithA(std::shared_ptr<A>&)
11:7: note:   no known conversion for argument 1 from 'std::shared_ptr<B>' to 'std::shared_ptr<A>&'

The reasoning behind the code is that there might be many derived classes and I know only at runtime which one is to be instantiated. The class C would decide to assign one of many derived types depending on what happens at runtime.

I didn't want to create several overloads of doSomethingWithA() and in general it is legal to assign a shared_ptr of a derived class to a shared_ptr of a base class, so I don't understand why the conversion is not possible.

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萌化 2025-01-21 00:41:52

您可以使用 std::static_pointer_cast 在这些类型之间进行转换，例如

#include <memory>

class A {
};

class B : public A {
};

class C {
public:
    void doSomethingWithA(std::shared_ptr<A>& a) {
        a = std::make_shared<B>();
    }

};

int main()
{
    std::shared_ptr<B> b;
    C c;
    std::shared_ptr<A> a_ptr_from_b = std::static_pointer_cast<A>(b);
    c.doSomethingWithA(a_ptr_from_b);
}

如果您需要的话，还有一个 std::dynamic_pointer_cast 但在本例中是 static会起作用的。

You can convert between these types with std::static_pointer_cast e.g.

#include <memory>

class A {
};

class B : public A {
};

class C {
public:
    void doSomethingWithA(std::shared_ptr<A>& a) {
        a = std::make_shared<B>();
    }

};

int main()
{
    std::shared_ptr<B> b;
    C c;
    std::shared_ptr<A> a_ptr_from_b = std::static_pointer_cast<A>(b);
    c.doSomethingWithA(a_ptr_from_b);
}

There's also an std::dynamic_pointer_cast if you ever need that but in this case static will work.

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