Question

class A {};

typedef shared_ptr<const A*> AConstPtr;
typedef shared_ptr<A*> APtr;

vector<APtr> ptr;

const vector<AConstPtr>* foo()
{
    return &ptr;
}

This code does not compile, because "there is no implicit conversion from vector<Aptr>* to const vector<AConstPtr> * " Is there anyway to make this work, without creating a new vector, and without using an unsafe cast?

因此,我之所以需要,是因为我有一个在内部把清单作为矢量代码(<><Ptr>的班子,但需要通过接口使清单完全一致。

Answer 1

There s no way to do such a conversion since different shared_ptrs aren t related types.

First, are you really sure that you need to expose the implementation detail that there s an internal vector of shared pointers? That s really going to tie you to that implementation and it won t be subject to change without breaking API.

如何使用@Cubbi的建议,以及用<条码>取代<条码>的接口? 然后,你可以很容易地向外部客户代表一个集装箱,而不把 yourself打到<条码>>。

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