Can I implicitly create a trivially copiable type

问题

My question is this:

Suppose type T is trivially copyable....can "create" an instance of this type without calling a constructor....like so:

#include <type_traits>
#include <cstring>

using T = int; // T can be any trivially copyable type


T create(const T& other)
{
    std::aligned_storage_t<sizeof(T),alignof(T)> my_T;

    std::memcpy(&my_T,&other,sizeof(T));

    return *reinterpret_cast<T*>(&my_T);
}

Is this defined behavior, or can I only copy into an existing object of type T?

回答1:

The rule, from [intro.object], is:

An object is created by a definition ([basic.def]), by a new-expression, when implicitly changing the active member of a union ([class.union]), or when a temporary object is created ([conv.rval], [class.temporary]).

None of those things is happening here, so you don't have an object. There is implicit object creation. That cast is technically UB because there is no object of type T at my_T.

What trivially copyable means is that if you then copied the bytes from my_T into another T object, then you would get the same behavior as if you'd just copied the T using the copy constructor. But you still need the objects to have already existed.

---

Note that this is an area that is being actively worked on, via P0593. There are many, many places where you really just need to say "I have a T here" and have that work, and every compiler already allows for that. It's just that that concept exists far outside of the object model that we have in C++ today.

来源：https://stackoverflow.com/questions/51572788/can-i-implicitly-create-a-trivially-copiable-type