Why are recursive struct types illegal in Rust?

Question

I'm trying out random things to deepen my understanding of Rust. I just ran into the following error with this code:

struct Person {
    mother: Option<Person>,
    father: Option<Person>,
    partner: Option<Person>,
}

pub fn main() {
    let susan = Person {
        mother: None,
        father: None,
        partner: None,
    };

    let john = Person {
        mother: None,
        father: None,
        partner: Some(susan),
    };
}

The error is:

error[E0072]: recursive type `Person` has infinite size
 --> src/main.rs:1:1
  |
1 | struct Person {
  | ^^^^^^^^^^^^^ recursive type has infinite size
2 |     mother: Option<Person>,
  |     ---------------------- recursive without indirection
3 |     father: Option<Person>,
  |     ---------------------- recursive without indirection
4 |     partner: Option<Person>,
  |     ----------------------- recursive without indirection
  |
  = help: insert indirection (e.g., a `Box`, `Rc`, or `&`) at some point to make `Person` representable

I understand that I can fix it if I put the Person in a Box, so this works:

struct Person {
    mother: Option<Box<Person>>,
    father: Option<Box<Person>>,
    partner: Option<Box<Person>>,
}

pub fn main() {
    let susan = Person {
        mother: None,
        father: None,
        partner: None,
    };

    let john = Person {
        mother: None,
        father: None,
        partner: Some(Box::new(susan)),
    };
}

I would like to understand the full story behind that. I know that boxing means that it will be stored on the heap rather than the stack but I don't get why this indirection is necessary.

Answer 1

Data inside structs and enums (and tuples) is stored directly inline inside the memory of the struct value. Given a struct like

struct Recursive {
    x: u8,
    y: Option<Recursive>
}

let's compute the size: size_of::<Recursive>(). Clearly it has 1 byte from the x field, and then the Option has size 1 (for the discriminant) + size_of::<Recursive>() (for the contained data), so, in summary, the size is the sum:

size_of::<Recursive>() == 2 + size_of::<Recursive>()

That is, the size would have to be infinite.

Another way to look at it is just expanding Recursive repeatedly (as tuples, for clarity):

Recursive ==
(u8, Option<Recursive>) ==
(u8, Option<(u8, Option<Recursive>)>) ==
(u8, Option<(u8, Option<(u8, Option<Recursive>)>)>) ==
...

and all of this is stored inline in a single chunk of memory.

A Box<T> is a pointer, i.e. it has a fixed size, so (u8, Option<Box<Recursive>>) is 1 + 8 bytes. (One way to regard Box<T> is that it's a normal T with the guarantee that it has a fixed size.)

Answer 2

The Rust Programming Language has this to say about recursive types:

Rust needs to know at compile time how much space a type takes up. One kind of type whose size can’t be known at compile time is a recursive type where a value can have as part of itself another value of the same type. This nesting of values could theoretically continue infinitely, so Rust doesn’t know how much space a value of a recursive type needs. Boxes have a known size, however, so by inserting a box in a recursive type definition, we are allowed to have recursive types.

Basically, the struct would be of infinite size if you don't use boxing. E.g., Susan has a mother, father, and partner, each of which have a mother, father, and partner....etc. Boxing uses a pointer, which is a fixed size, and dynamic memory allocation.