A shared reference, &T is , as the name implies, a pointer that may be shared. Any number of references may exist to the same value, and each shared reference is Copy, so you can trivially make more of them

I don't understand why a shared reference has to implement copy. In fact, isn't this not true just by the fact that references work for Strings and Strings size can't be known at compile time?

I'm having trouble with the idea of assigning a new value to a mutable reference.

let mut x = Box::new(42); *x = 84;

Why in this example, is the assignment dereferenced. Why not just do x=84? is it dereferenced specifically because is Boxed on the heap?

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[–] BatmanAoD 13 points 1 year ago (1 children)

The key thing to understand is that in Rust, references are considered unique types. This means that &T is a separate type from T.

So, for #1, it is not saying that T implements Copy, it is saying that regardless of what T is, &T implements Copy. This is because, by definition, it is always valid to copy a shared reference, even if T itself is not Copy.

Part of the reason this is confusing is that traits often include references in their function signatures; and in particular, Clone::clone has the signature fn clone(&self) -> Self. So when T implements clone, it has a method that takes &T and returns T. But even though the signature takes &T, the type that implements Clone is T itself. (&T always implements Clone as well, just as it always implements Copy, but as with Copy, this is independent from whether T itself implements Clone. See for example the error message you get when explicitly cloning a shared reference: https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=a1b80cc83570321868c4ad55ee3353dc)

Since Copy is a marker trait, it doesn't have any associated methods making it explicit that Copy affects how &T can be used. However, Copy requires the type to implement Clone (even though you can implement Clone in terms of Copy) and implies that T can always be automatically created from &T without an explicit call to T::clone, so you can think of the "signature" for Copy as matching that of Clone, even though there's no actual copy method.

For #2, I recommend thinking in terms of explicit types. Adding annotations, you get:

let mut x: Box = Box::new(42); *x = 84_i32;

The type of x is Box. You cannot assign an i32 to a Box; they're different types! But the type of *x is i32, and thus you can assign 84 to it.

The trait used to make Box behave this way is DerefMut, which explicitly makes *x assignable: https://doc.rust-lang.org/std/ops/trait.DerefMut.html

[–] nerdblood 2 points 1 year ago* (last edited 1 year ago) (1 children)

So, for #1, it is not saying that T implements Copy, it is saying that regardless of what T is, &T implements Copy. This is because, by definition, it is always valid to copy a shared reference, even if T itself is not Copy.

Got it! this helps a lot, thanks. I think I was indeed thinking of it as some kind of pseudo type and not a type in of itself. I got a little lost in the further explanation, but I really need to spend more time understanding Clone and Copy in general. Thanks!

[–] BatmanAoD 1 points 1 year ago

It's totally reasonable to assume that &T is a "pseudo-type"; that's very much what it is in C++, and there aren't any other mainstream languages with that syntax to compare against!