CODEWITHSUNDEEP
rust
Elf Sternberg
Elf Sternberg

Reputation: 16361

In Rust, how can a reference be a pointer to a pointer-to-a-pointer?

Today's Rust mystery is from section 4.9 of The Rust Programming Language, First Edition. The example of references and borrowing has this example:

fn main() {
    fn sum_vec(v: &Vec<i32>) -> i32 {
        return v.iter().fold(0, |a, &b| a + b);
    }

    fn foo(v1: &Vec<i32>) -> i32 {
        sum_vec(v1);
    }

    let v1 = vec![1, 2, 3];

    let answer = foo(&v1);
    println!("{}", answer);
}

That seems reasonable. It prints "6", which is what you'd expect if the v of sum_vec is a C++ reference; it's just a name for a memory location, the vector v1 we defined in main().

Then I replaced the body of sum_vec with this:

fn sum_vec(v: &Vec<i32>) -> i32 {
    return (*v).iter().fold(0, |a, &b| a + b);
}

It compiled and worked as expected. Okay, that's not… entirely crazy. The compiler is trying to make my life easier, I get that. Confusing, something that I have to memorize as a specific tic of the language, but not entirely crazy. Then I tried:

fn sum_vec(v: &Vec<i32>) -> i32 {
    return (**v).iter().fold(0, |a, &b| a + b);
}

It still worked! What the hell?

fn sum_vec(v: &Vec<i32>) -> i32 {
    return (***v).iter().fold(0, |a, &b| a + b);
}

type [i32] cannot be dereferenced. Oh, thank god, something that makes sense. But I would have expected that almost two iterations earlier!

References in Rust aren't C++ "names for another place in memory," but what are they? They're not pointers either, and the rules about them seem to be either esoteric or highly ad-hoc. What is happening such that a reference, a pointer, and a pointer-to-a-pointer all work equally well here?

Upvotes: 10

Views: 1912

Answers (1)

Shepmaster
Shepmaster

Reputation: 430671

The rules are not ad-hoc nor really esoteric. Inspect the type of v and it's various dereferences:

fn sum_vec(v: &Vec<i32>) {
    let () = v;
}

You'll get:

  1. v -> &std::vec::Vec<i32>
  2. *v -> std::vec::Vec<i32>
  3. **v -> [i32]

The first dereference you already understand. The second dereference is thanks to the Deref trait. Vec<T> dereferences to [T].

When performing method lookup, there's a straight-forward set of rules:

  1. If the type has the method, use it and exit the lookup.
  2. If a reference to the type has the method, use it and exit the lookup.
  3. If the type can be dereferenced, do so, then return to step 1.
  4. Else the lookup fails.

References in Rust aren't C++ "names for another place in memory,"

They absolutely are names for a place in memory. In fact, they compile down to the same C / C++ pointer you know.

Upvotes: 10

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