问题
I was reading the documentation for std::any_cast and I find it strange that the API has the cast either return a value to the held object or a pointer to it. Why not return a reference? A copy needs to be made every time the function is called with a non pointer type argument.
I can see that the pointer version of the cast might signal intentions a bit more and might be a bit more clear but why not have the value returned be a reference like this?
template<typename ValueType>
ValueType& any_cast(any* operand);
instead of
template <typename ValueType>
ValueType* any_cast(any* operand);
Further it seems like even if you ask for a reference the cast removes the reference and returns a copy to the stored object see the explanations for the return values for function overloads 1-3 here http://en.cppreference.com/w/cpp/utility/any/any_cast
回答1:
You can see a discussion regarding the C++ standard for this here: https://groups.google.com/a/isocpp.org/forum/#!topic/std-proposals/ngSIHzM6kDQ
Note that Boost has defined any_cast this way for more than a decade, plus it matches static_cast and friends. So if you want a reference, do this:
any_cast<Foo&>(x)
The same as you'd do for the older _casts in C++.
来源：https://stackoverflow.com/questions/41905868/stdany-cast-returns-a-copy