Rust：引用如何成为类型？

Question

So what I am asking is, what is the difference between the return types, &std::vec::Vec and std::vec::Vec? Just curious. If I make a reference to something, I'm not creating a new type. It still retains its data and structure and so retains its type. But for some reason I get this error:

   error[E0308]: mismatched types
   --> src/cam.rs:170:3
    |
168 |     pub fn index2d(self, x: usize, y: usize) -> Vec<u8> {
    |                                                 ------- expected `std::vec::Vec<u8>` because of return type
169 |         let c = &self.pyxels[y*WIDTH+x];
170 |         c
    |         ^
    |         |
    |         expected struct `std::vec::Vec`, found reference
    |         help: try using a conversion method: `c.to_vec()`
    |
    = note: expected type `std::vec::Vec<u8>`
               found type `&std::vec::Vec<u8>`
                           ^ (umm excuse me?)

That one little symbol (&) really seems to make all the difference and I have no clue why.

Answer 1

If I make a reference to something, I'm not creating a new type.

If by "creating a new type" , you mean "creating an object of a different type" , then yes, that's exactly what you're doing. A reference to a thing is not that thing. It's similar to the difference between having a house, and having a slip of paper with the address of a house written on it.

Though the syntax of Rust often makes access to an object through a reference look identical to direct access to that object.

let v: Vec<u8> = Vec::new();
let vr = &v;

assert_eq!(v.len(), rv.len());

So perhaps that is why you are confused?