std :: shared_ptr向上转换为基类-最佳方法? [英] std::shared_ptr upcasting to base class - best method?

问题描述

哪种转换效果更好，有什么区别?

Which conversion is better, and what is the difference?

class Base
{};

class Derived : public Base, public std::enable_shared_from_this<Derived>
{};

int main(int argc, const char * argv[])
{
    std::shared_ptr<Base> ptr1 = std::dynamic_pointer_cast<Base>(std::shared_ptr<Derived>(new Derived())); // version 1
    std::shared_ptr<Base> ptr2 = std::shared_ptr<Derived>(new Derived()); // version 2
    return 0;
}

推荐答案

与在 shared_ptr 的其他用例中一样，您应该更喜欢使用 make_shared 而不是构造 shared_ptr :

As in other use cases of shared_ptr, you should prefer using make_shared instead of constructing the shared_ptr manually:

std::shared_ptr<Base> ptr2 = std::make_shared<Derived>();

这实际上是您的版本2，再加上 make_shared的各种好处 .

This is essentially your version 2, plus the various benefits of make_shared.

版本1做了很多不必要的工作:首先，您构造一个临时的 shared_ptr< Derived> ，然后您将 dynamic_cast 的内容指向基类指针(而 static_cast 就足够了)，然后将该指针存储在另一个 shared_ptr< Base> 中.因此，您有许多不必要的运行时操作，但与版本2相比，类型安全性没有任何好处.

Version 1 does a bunch of unnecessary stuff: First you construct a temporary shared_ptr<Derived>, then you dynamic_cast its contents to a base class pointer (while a static_cast would be sufficient here) and then you store that pointer in a different shared_ptr<Base>. So you have a lot of unnecessary runtime operations, but no benefits regarding type safety over Version 2.

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