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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[–] KillTheMule 9 points 1 year ago* (last edited 1 year ago) (2 children)

A reference IS Copy, by the simple fact that it is a primitive value on the stack.

This seems a bit misleading, noting that unique/mutable references aren't Copy. Shared references are Copy because it's sound to have that, and it's a huge QOL improvement over the alternative.

[–] [email protected] 2 points 1 year ago

In this context mutability is part of the type signatures. &T and &mut T are two different types, the former implements copy but not the later. It's not really an "exception" in the type system.

[–] Anders429 1 points 1 year ago (1 children)

I wouldn't say it's misleading. The question was specifically about shared references, it seemed obvious to me that's what they were referring to in their answer.

[–] KillTheMule 2 points 1 year ago

The question was specifically about shared references

Sure, but the way I read the answer was "All primitive values on the stack are Copy", which isn't true (my example being mutable references, which have the same representation as shared ones, "just" a different semantic meaning). That's what I meant by misleading.