Why Reference a Reference

[ocu]

Why

let x = 1;
let y = &x;
let z = y;

Instead of

let x = 1;
let y = &x;
let z = &x;

[jofas]

You might want to provide a bit more context. Both snippets are semantically equivalent, shared references have copy semantics (z does not "reference a reference" as from your title, but copies the reference to x from y).

[ocu]

So

let z = y;

is equivalent to

let z = &y.clone();

and

let z = &x;

but not to

let z = &y;

[jofas]

@ocu

let z = &y;

Yes, that is exactly how you can reference y. z will have the type &&i32 in this case.

[ocu]

I see. What would the use cases for &&type references be?

[jofas]

You sometimes have to deal with &&T in call-back based APIs, like when you are iterating over an Iterator<Item=&T> and use Iterator::find, for example, which takes a callback with &Item as its argument. Otherwise I don't recall having to deal with references-to-references very often. It is generally not a very useful construct, as references are trivially copyable.

[ocu]

Thank you very much, this has all been really helpful.