英文:
Why can I bind base class shared_ptr rvalue reference to derived class shared_ptr?
问题
我知道我们不能将右值引用绑定到左值，所以下面的代码不会编译：
class Base {};
int main(int argc, char** argv) {
std::shared_ptr<Base> base = std::make_shared<Base>();
std::shared_ptr<Base>&& ref = base;
}
但是下面的代码可以编译，我搞不清楚为什么。
class Base {};
class Derived : public Base {};
int main(int argc, char** argv) {
std::shared_ptr<Derived> derived = std::make_shared<Derived>();
std::shared_ptr<Base>&& ref = derived;
}
使用类型推断，auto&& 和 T&& 都是“正向引用（forward reference）”，因为它们可以是左值引用、常量引用或右值引用。但在我的情况下（或者我认为的情况下），没有类型推断，也许在 shared_ptr 中有，但我引用的是 shared_ptr 本身，而不是它的底层对象，所以我困惑了。
英文:
so I know we cannot bind rvalue reference to an lvalue, so the following code won't compile:
class Base {};
int main(int argc, char** argv) {
std::shared_ptr<Base> base = std::make_shared<Base>();
std::shared_ptr<Base>&& ref = base;
}
but the next code can compile, and I can't figure out why.
class Base {};
class Derived : public Base {};
int main(int argc, char** argv) {
std::shared_ptr<Derived> derived = std::make_shared<Derived>();
std::shared_ptr<Base>&& ref = derived;
}
With type deduction, auto&& and T&& are forward reference, because they can be lvaue reference, const reference or rvalue reference.
But here is no type deduction in my case(or what I think), maybe in share_ptr, but I'm referencing share_ptr itself, not its underlying object, so I'm stuck.
答案1
得分: 6
std::shared_ptr<Derived> 隐式可转换为 std::shared_ptr<Base>。当这种情况发生时，会创建一个新的临时对象，可以绑定到 std::shared_ptr<Base>&&。由于该对象是一个 prvalue，其生命周期会延长到引用的生命周期。
英文:
std::shared_ptr<Derived> is implicitly convertible to std::shared_ptr<Base>. When that happens, a new, temporary object is created, which a std::shared_ptr<Base>&& can bind to. Since that object is a prvalue, its lifetime gets extended to the lifetime of the reference.
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