- Feature Name: N/A
- Start Date: 01 March, 2016
- RFC PR: rust-lang/rfcs#1521
- Rust Issue: rust-lang/rust#33416
With specialization on the way, we need to talk about the semantics of
<T as Clone>::clone() where T: Copy.
It's generally been an unspoken rule of Rust that a
clone of a
Copy type is
equivalent to a
memcpy of that type; however, that fact is not documented
anywhere. This fact should be in the documentation for the
Clone trait, just
like the fact that
T: Eq should implement
a == b == c == a rules.
Vec::clone() is implemented by creating a new
Vec, and then
cloning all of the elements from one into the other. This is slow in debug mode,
and may not always be optimized (although it often will be). Specialization
would allow us to simply
memcpy the values from the old
Vec to the new
Vec in the case of
T: Copy. However, if we don't specify this, we will not
be able to, and we will be stuck looping over every value.
It's always been the intention that
Clone::clone == ptr::read for T: Copy; see
issue #23790: "It really makes sense for
Clone to be a
Copy is a refinement of
suffices, basically." This idea was also implicit in accepting
rfc #0839 where "[B]ecause Copy: Clone, it would be backwards
compatible to upgrade to Clone in the future if demand is high enough."
<T as Clone>::clone(t) shall be equivalent to
T: Copy, t: &T. An implementation that does not uphold this shall not
result in undefined behavior;
Clone is not an
Also add something like the following sentence to the documentation for the
x: T, and
y: &T, then
let x = y.clone(); is equivalent to
let x = *y;. Manual implementations must be careful to uphold this."
This is a breaking change, technically, although it breaks code that was malformed in the first place.
The alternative is that, for each type and function we would like to specialize
in this way, we document this separately. This is how we started off with
What the exact wording should be.