My question is the following:
If I want to copy a class type, memcpy can do it very fast. This is allowed in some situations.
We have some type traits:
What I would like to know is the exact requirements when a type will be "bitwise copyable".
My conclusion is that a type is bitwise copyable if both of is_trivally_copyable
and is_standard_layout
traits are true:
P.S.: of course, the result of memcpy must be correct. I know I could memcpy in any situation but incorrectly.
You can copy an object of type T using memcpy
when is_trivially_copyable<T>::value
is true. There is no particular need for the type to be a standard layout type. The definition of 'trivially copyable' is essentially that it's safe to do this.
An example of a class that is safe to copy with memcpy
but which is not standard layout:
struct T { int i; private: int j; };
Because this class uses different access control for different non-static data members it is not standard layout, but it is still trivially copyable.
If is_trivally_copyable<T>::value
(or in C++14 is_trivially_copyable<T>()
, or in C++17 is_trivially_copyable_v<T>
) is not zero, the type is copyable using memcpy
.
Per the C++ standard, a type being trivially copyable means:
the underlying bytes making up the object can be copied into an array of char or unsigned char. If the content of the array of char or unsigned char is copied back into the object, the object shall subsequently hold its original value.
However, it is important to realise that pointers are trivially copyable types, too. Whenever there are pointer inside the data structures you will be copying, you have to brainually make sure that copying them around is proper.
Examples where hazard may be caused by just relying on the object being trivially copyable:
So whenever memcopying, keep in mind to check whether pointers could be copied in that specific case, and if that would be okay.
Realise that is_trivially_copyable
is only the "Syntax Check", not the "Semantic Test", in compiler parlance.
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