I can agree that the example function is not the best usecase. But the point still stand that there’s no realistic escape hatch from lifetimes and memory management in Rust.
Cow does not work when you are actually required to return a reference, e.g. if you’re working with some other crate that requires that. Cow also has some more strict requirements on reborrows (i.e. you can reborrow a &'short &'long T to a &'long T, but you can only reborrow a &'short Cow<'long, T> to a &'short T).
LazyLock can solve very specific issues like static, but is not a general escape hatch. Again, the example is not the best to showcase this, but imagine if you have to perform this operation for an unknown amount of runtime values. LazyLock will only work for the very first one.
Option
not an option?Cow
s?LazyLock
static?I can agree that the example function is not the best usecase. But the point still stand that there’s no realistic escape hatch from lifetimes and memory management in Rust.
Cow
does not work when you are actually required to return a reference, e.g. if you’re working with some other crate that requires that.Cow
also has some more strict requirements on reborrows (i.e. you can reborrow a&'short &'long T
to a&'long T
, but you can only reborrow a&'short Cow<'long, T>
to a&'short T
).LazyLock
can solve very specific issues likestatic
, but is not a general escape hatch. Again, the example is not the best to showcase this, but imagine if you have to perform this operation for an unknown amount of runtime values.LazyLock
will only work for the very first one.What does that even mean? Can you provide a practical example?
(I’m assuming you’re familiar with
Deref
and autoref/autoderef behaviors in Rust.)