返回shared_ptr时如何完成协变返回类型? [英] How to accomplish covariant return types when returning a shared_ptr?

问题描述

using namespace boost;

class A {};
class B : public A {};

class X {
  virtual shared_ptr<A> foo();
};

class Y : public X {
  virtual shared_ptr<B> foo();
};

返回类型不是协变的(因此也不合法)，但是如果我使用的是原始指针，它们将是协变的.解决这个问题的普遍接受的成语是什么?

The return types aren't covariant (nor are they, therefore, legal), but they would be if I was using raw pointers instead. What's the commonly accepted idiom to work around this, if there is one?

推荐答案

我认为解决方案根本上是不可能的，因为协方差取决于与智能指针不兼容的指针算法.

I think that a solution is fundamentally impossible because covariance depends on pointer arithmetic which is incompatible with smart pointers.

当Y::foo将shared_ptr<B>返回给动态调用者时，必须在使用前将其强制转换为shared_ptr<A>.就您而言，B*可以(可能)被简单地重新解释为A*，但是对于多重继承，您将需要一些魔术来向C ++讲述static_cast<A*>(shared_ptr<B>::get()).

When Y::foo returns shared_ptr<B> to a dynamic caller, it must be cast to shared_ptr<A> before use. In your case, a B* can (probably) simply be reinterpreted as an A*, but for multiple inheritance, you would need some magic to tell C++ about static_cast<A*>(shared_ptr<B>::get()).

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