问题描述
我想在某个时候使用 boost::log，但我不能将 std::shared_ptr 作为参数传递，因为编译器 (VS2010) 无法转换将其转换为 boost::shared_ptr.
I want to use boost::log at some point, but I cannot pass a std::shared_ptr as a parameter, because the compiler (VS2010) cannot convert it into a boost::shared_ptr.
我真的不喜欢他们彼此陌生的事实.
I don't really like the fact that they are aliens to one another.
是否有一种安全透明的方式将一种转换为另一种，以免它们相互绊倒?
Is there a safe and transparent way to convert one into the another, so as they don't stumble over each other?
我不认为它与这个问题重复表示两者是相同的.
I don't think it is duplicate of this question that states both are the same.
推荐答案
你可以这样做:
template<typename T>
boost::shared_ptr<T> make_shared_ptr(std::shared_ptr<T>& ptr)
{
return boost::shared_ptr<T>(ptr.get(), [ptr](T*) mutable {ptr.reset();});
}
template<typename T>
std::shared_ptr<T> make_shared_ptr(boost::shared_ptr<T>& ptr)
{
return std::shared_ptr<T>(ptr.get(), [ptr](T*) mutable {ptr.reset();});
}
请注意，这不适用于对源 ptr 的弱引用.所以要小心那些！
Note that this does not work with weak references to the source ptr. So be careful with those!
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