即使在 C++20 中 std::memcpy 也不是 constexpr 的原因是什么？

Question

I understand that copying arbitrary chunks of memory is not always possible to do at compile time but since we are getting constexpr constainers, virtual methods and also algorithms, why not memcpy too? It is too a kind of algorithm.

Furthemore,

• C++20 std::bit_cast seems a lot like std::memcpy workaround reinterpret_cast but it is constexpr .
• std::copy using iterators is marked as constexpr for C++20, so copying is somehow possible for types.

The usage would be to either copy or just "reinterpret" variables/arrays in constexpr functions, the former is not solved by std::bit_cast AFAIK. In particular, the question and my answer would like to use it.

• Is there any particular reason for why std::bit_cast can be constexpr but std::memcpy cannot?
• Does it have to do with memcpy using void pointers instead of typed references?
• Not actually having to copy anything?
• C backwards compatibility?
• Maybe because there is no support for a "pointer to constexpr memory"? But the same applies to the reference parameter in std::bit_cast and iterators in std::copy .

Relevant answer to C++20 bit_cast vs reinterpret_cast briefly cites from somewhere:

Furthermore, it is currently impossible to implement a constexpr bit-cast function, as memcpy itself isn't constexpr. Marking the proposed function as constexpr doesn't require or prevent memcpy from becoming constexpr, but requires compiler support. This leaves implementations free to use their own internal solution (eg LLVM has a bitcast opcode).

But it does not go into detail of not making it constexpr too.

Note, that I do not ask for why std::bit_cast exists. I like it, it provides a clear intention instead of std::memcpy workaround.

Answer 1

The C++ object model in runtime code is generally treated somewhat loosely. It has fairly strict rules, but there are a bunch of backdoors that are either allowed or declared UB. The latter means that you can still write code to do it, but C++ guarantees nothing about the behavior of that code.

Within constant evaluation (aka: compile-time execution of code), this is not the case. The restrictions on constexpr are specifically intended to allow the object model to be a real thing that you must follow, with no viable backdoors. And even the ones that it occasionally permits are explicitly required to be ill-formed and produce a compile-error, rather than being silent UB.

Basically at runtime, you get to treat memory as just bytes of storage. At compile-time, you can't ; you're not allowed to. Even with dynamic allocation in constexpr code added in C++20, you don't get to play a lot of the games you usually get to play with that sort of thing.

memcpy deals in bytes of storage, copying them back and forth with no idea what they mean. bitcast knows both the source and destination objects, and it will not allow you to do it unless the source and destination objects are appropriate for bitcast ing (ie: trivially-copyable).

bitcast also has very specific restrictions on the content of both such objects if you want it to work at compile-time. In particular, you can't bitcast pointers or any objects containing pointers of any kind. Or references.

This is because pointers at compile-time are not just addresses. In order to catch UB, a compile-time pointer has to know the true dynamic type of the object it points to. So pointer conversions that just convert the address aren't allowed at compile-time.

Answer 2

That's more a comment then an answer as I'm only citing what is written in P0202R0: Add Constexpr Modifiers to Functions in and Headers , but I write it here as is does not fit the comments:

B. std::memmove and std::memcpy must have constexpr additions
std::memmove and std::memcpy accept void* and const void* parameters. This makes them impossible to implement in pure C++ as constexpr , because constant expressions can not evaluate a conversion from type cv void * to a pointer-to-object type according to [expr.const].
However those functions are not only popular, but also are widely used across Standard Library to gain better performance. Not making them constexpr will force standard Library developer to have compiler intrinsics for them anyway. This is a hard step that must be done.

The related section of [expr.const] :

8.6 Constant expressions [expr.const]
[…]An expression e is a core constant expression unless the evaluation of e , following the rules of the abstract machine (6.8.1), would evaluate one of the following expressions:
[…]
(2.13) — a conversion from type cv void* to a pointer-to-object type;