Despite thoroughly reading the documentation, I'm rather confused about the meaning of the &
and *
symbol in Rust, and more generally about what is a Rust reference exactly.
In this example, it seems to be similar to a C++ reference (that is, an address that is automatically dereferenced when used):
fn main() {
let c: i32 = 5;
let rc = &c;
let next = rc + 1;
println!("{}", next); // 6
}
However, the following code works exactly the same:
fn main() {
let c: i32 = 5;
let rc = &c;
let next = *rc + 1;
println!("{}", next); // 6
}
Using *
to dereference a reference wouldn't be correct in C++. So I'd like to understand why this is correct in Rust.
My understanding so far, is that, inserting *
in front of a Rust reference dereferences it, but the *
is implicitly inserted anyway so you don't need to add it (while in C++, it's implicitly inserted and if you insert it you get a compilation error).
However, something like this doesn't compile:
fn main() {
let mut c: i32 = 5;
let mut next: i32 = 0;
{
let rc = &mut c;
next = rc + 1;
}
println!("{}", next);
}
error[E0369]: binary operation `+` cannot be applied to type `&mut i32`
--> src/main.rs:6:16
|
6 | next = rc + 1;
| ^^^^^^
|
= note: this is a reference to a type that `+` can be applied to; you need to dereference this variable once for this operation to work
= note: an implementation of `std::ops::Add` might be missing for `&mut i32`
But this works:
fn main() {
let mut c: i32 = 5;
let mut next: i32 = 0;
{
let rc = &mut c;
next = *rc + 1;
}
println!("{}", next); // 6
}
It seems that implicit dereferencing (a la C++) is correct for immutable references, but not for mutable references. Why is this?
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