shared_ptr 和切片 [英] shared_ptr and slicing

问题描述

与我共事的人曾经说过 shared_ptr 是不安全的，并且会在从派生类转换到基类(即向上转换)时进行切片.例如，如果有 2 个类 A 和 B，其中 B 派生自 A，则

Someone I worked with once said that shared_ptr was unsafe and would slice when casting from a derived class to a base class (i.e. upcasting). For example if there were 2 classes A and B where B derived from A, then

shared_ptr<A> a(new B)

会切片.我向他指出 http://www.boost.org/doc/libs/1_43_0/libs/smart_ptr/shared_ptr.htm它说的地方

would slice. I pointed him to http://www.boost.org/doc/libs/1_43_0/libs/smart_ptr/shared_ptr.htm where it says

shared_ptr 可以隐式转换为 shared_ptr 只要 T* 可以隐式转换为 U*.

shared_ptr<T> can be implicitly converted to shared_ptr<U> whenever T* can be implicitly converted to U*.

暗示在这些情况下使用是安全的，但他似乎并不这么认为.

implied that it's safe to use in those contexts but he didn't seem to think so.

推荐答案

有人错了，对象切片 不适用于指针.指针用法被包裹在 shared_ptr 中并没有改变这一点 - 它在这里没有做任何特殊的魔法，它使用传递给其构造函数的值初始化一个内部指针.

That someone is wrong, object slicing doesn't apply to pointers. That the pointer usage is wrapped away in a shared_ptr doesn't change that - it doesn't do any particular magic here, it initializes an internal pointer with the value passed to its constructor.

简化它可以看起来例如出于这个问题的目的，像这样:

Simplified it could look e.g. like this for the purpose of this question:

template<class T> struct ptr {
    T* t;
    ptr(T* t) : t(t) {}
    // ...
};

您丢失了 B 的静态类型信息，是的，但是指向的对象没有任何变化.

You lose the static type-information of B, yes, but nothing changes regarding the object pointed to.

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