In this example, you could have just made a constant with value 0 and returned a reference to that. It would also have a 'static lifetime and there would be no leaking.
I believe the intention was to demonstrate something that works with runtime values too, and a constant 0 does not.
Btw you can just write &0 to do what you proposed, there’s no need for an explicit constant/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 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.
(Note that I’m not the article author)
I believe the intention was to demonstrate something that works with runtime values too, and a constant 0 does not.
Btw you can just write
&0to do what you proposed, there’s no need for an explicit constant/static.Optionnot an option?Cows?LazyLockstatic?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.
Cowdoes not work when you are actually required to return a reference, e.g. if you’re working with some other crate that requires that.Cowalso has some more strict requirements on reborrows (i.e. you can reborrow a&'short &'long Tto a&'long T, but you can only reborrow a&'short Cow<'long, T>to a&'short T).LazyLockcan 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.LazyLockwill 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
Derefand autoref/autoderef behaviors in Rust.)