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impl Clone for Cow #19359

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Dec 5, 2014
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impl Clone for Cow #19359

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12 changes: 12 additions & 0 deletions src/libcore/borrow.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -137,6 +137,18 @@ pub enum Cow<'a, T, Sized? B: 'a> where B: ToOwned<T> {
Owned(T)
}

impl<'a, T, Sized? B> Clone for Cow<'a, T, B> where B: ToOwned<T> {
fn clone(&self) -> Cow<'a, T, B> {
match *self {
Borrowed(b) => Borrowed(b),
Owned(ref o) => {
let b: &B = BorrowFrom::borrow_from(o);
Owned(b.to_owned())
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I must confess that it feels weird to me to always into_owned a Clone of an owned Cow. I imagine that always producing a Borrow would have unfortunate consequences? Can you provide an example?

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If Self = SendStr = CowString<'static>. Cloning that can't return a Borrowed value, since that would need an inner &'static str, and you can't produce &'static str from String.

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Should we explicitly document the behaviour, since it's subtle and might be unexpected? Or do you think it can be considered an implementation detail?

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Another option would be to require T: Clone instead of B: BorrowFrom<T> so you could just call .clone() on the owned variant (which may be more expected)

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require T: Clone instead of B: BorrowFrom

Note that the Cow struct enforces the B: BorrowFrom<T> bound, so it can't be removed. I wrote the impl in this way to reduce the numbers of bounds on the input parameters.

},
}
}
}

impl<'a, T, Sized? B> Cow<'a, T, B> where B: ToOwned<T> {
/// Acquire a mutable reference to the owned form of the data.
///
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