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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[–] snaggen 11 points 1 year ago* (last edited 1 year ago) (6 children)

A reference &T holds a pointer, ie. the memory adress to the actual content of T

So, in the example x doesn't hold the value 42, it holds the memory adress to the memory there the integer value 42 is stored. So, to access the value, you need to dereference the reference. Which is why you need to use *x when you assign the value.

[–] snaggen 6 points 1 year ago (5 children)

And the Copy question.It is not that s reference has to implement Copy. A reference IS Copy, by the simple fact that it is a primitive value on the stack.

[–] nerdblood 1 points 1 year ago

I think where I was getting hung up here as I was still thinking of the value of the reference rather than the reference itself. That's why I brought up Strings, it didn't make sense that String would automatically implement Copy since they go on the heap. But I see now we're talking about the pointer reference TO that value.

Thanks!

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