为什么 shared_ptr 计数器在作为 const 引用传递给基 class 时递增?

Why is shared_ptr counter incremented when passed as const reference to base class?

为什么当我将它传递给需要 const shared_ptr<base>& 的函数时,shared_ptr<drived> 计数器递增?

this question其中一个答案中提到:

shared_ptr<Base> and shared_ptr<Derived> are not covariant

我怀疑这与我的问题有关。它们不协变是什么意思?

这里是展示场景的代码片段:

#include <iostream>
#include <memory>

class Base {};

class Derived : public Base {};

void f(const std::shared_ptr<Base>& x)
{
    std::cout << "in function expecting const shared_ptr<Base>& - Use count: " << x.use_count() << std::endl;
}

int main(int argc, char const *argv[])
{
    std::cout << "Base class" << std::endl;
    auto a = std::make_shared<Base>();
    std::cout << "Created shared_ptr:  Initial use count: " << a.use_count() << std::endl;
    f(a);

    std::cout << "------------------\nChild class" << std::endl;
    auto b = std::make_shared<Derived>();
    std::cout << "Created shared_ptr. Initial use count: " << b.use_count() << std::endl;
    f(b);

    return 0;
}

结果:

>> g++ -std=c++17 -O2 -Wall -pedantic -pthread main.cpp && ./a.out
Base class
Created shared_ptr:  Initial use count: 1
in function expecting const shared_ptr<Base>& - Use count: 1
------------------
Child class
Created shared_ptr. Initial use count: 1
in function expecting const shared_ptr<Base>& - Use count: 2

shared_ptr<Derived> 不是 shared_ptr<Base>。他们是完全不同的类型。

为了从 shared_ptr<Derived> 中获得 shared_ptr<Base>,您需要创建一个。编译器可以调用 constructor 因为它没有被显式标记。这将增加使用次数,因为他们共享所有权。

template< class Y > shared_ptr( const shared_ptr<Y>& r ) noexcept;

Constructs a shared_ptr which shares ownership of the object managed by r. If r manages no object, *this manages no object too. The template overload doesn't participate in overload resolution if Y* is not implicitly convertible to (until C++17) compatible with (since C++17) T*.

您可以自己看到,通过将 f() 更改为采用 non-const 引用,创建了一个新的 shared_ptr。编译器应该给你一个错误,因为你不能将临时绑定到 non-const 引用。 See here