Subtyping and Variance -

# Subtyping and Variance

Subtyping is a relationship between types that allows statically typed languages to be a bit more flexible and permissive.

The most common and easy to understand example of this can be found in languages with inheritance. Consider an Animal type which has an `eat()` method, and a Cat type which extends Animal, adding a `meow()` method. Without subtyping, if someone were to write a `feed(Animal)` function, they wouldn't be able to pass a Cat to this function, because a Cat isn't exactly an Animal. But being able to pass a Cat where an Animal is expected seems fairly reasonable. After all, a Cat is just an Animal and more. Something having extra features that can be ignored shouldn't be any impediment to using it!

This is exactly what subtyping lets us do. Because a Cat is an Animal and more we say that Cat is a subtype of Animal. We then say that anywhere a value of a certain type is expected, a value with a subtype can also be supplied. Ok actually it's a lot more complicated and subtle than that, but that's the basic intuition that gets you by in 99% of the cases. We'll cover why it's only 99% later in this section.

Although Rust doesn't have any notion of structural inheritance, it does include subtyping. In Rust, subtyping derives entirely from lifetimes. Since lifetimes are regions of code, we can partially order them based on the contains (outlives) relationship.

Subtyping on lifetimes is in terms of that relationship: if `'big: 'small` ("big contains small" or "big outlives small"), then `'big` is a subtype of `'small`. This is a large source of confusion, because it seems backwards to many: the bigger region is a subtype of the smaller region. But it makes sense if you consider our Animal example: Cat is an Animal and more, just as `'big` is `'small` and more.

Put another way, if someone wants a reference that lives for `'small`, usually what they actually mean is that they want a reference that lives for at least `'small`. They don't actually care if the lifetimes match exactly. For this reason `'static`, the forever lifetime, is a subtype of every lifetime.

Higher-ranked lifetimes are also subtypes of every concrete lifetime. This is because taking an arbitrary lifetime is strictly more general than taking a specific one.

(The typed-ness of lifetimes is a fairly arbitrary construct that some disagree with. However it simplifies our analysis to treat lifetimes and types uniformly.)

However you can't write a function that takes a value of type `'a`! Lifetimes are always just part of another type, so we need a way of handling that. To handle it, we need to talk about variance.

# Variance

Variance is where things get a bit complicated.

Variance is a property that type constructors have with respect to their arguments. A type constructor in Rust is a generic type with unbound arguments. For instance `Vec` is a type constructor that takes a `T` and returns a `Vec<T>`. `&` and `&mut` are type constructors that take two inputs: a lifetime, and a type to point to.

A type constructor F's variance is how the subtyping of its inputs affects the subtyping of its outputs. There are three kinds of variance in Rust:

- F is covariant over `T` if `T` being a subtype of `U` implies `F<T>` is a subtype of `F<U>` (subtyping "passes through")
- F is contravariant over `T` if `T` being a subtype of `U` implies `F<U>` is a subtype of `F<T>` (subtyping is "inverted")
- F is invariant over `T` otherwise (no subtyping relation can be derived)

It should be noted that covariance is far more common and important than contravariance in Rust. The existence of contravariance in Rust can mostly be ignored.

Some important variances (which we will explain in detail below):

- `&'a T` is covariant over `'a` and `T` (as is `*const T` by metaphor)
- `&'a mut T` is covariant over `'a` but invariant over `T`
- `fn(T) -> U` is contravariant over `T`, but covariant over `U`
- `Box`, `Vec`, and all other collections are covariant over the types of their contents
- `UnsafeCell<T>`, `Cell<T>`, `RefCell<T>`, `Mutex<T>` and all other interior mutability types are invariant over T (as is `*mut T` by metaphor)

To understand why these variances are correct and desirable, we will consider several examples.

We have already covered why `&'a T` should be covariant over `'a` when introducing subtyping: it's desirable to be able to pass longer-lived things where shorter-lived things are needed.

Similar reasoning applies to why it should be covariant over T: it's reasonable to be able to pass `&&'static str` where an `&&'a str` is expected. The additional level of indirection doesn't change the desire to be able to pass longer lived things where shorter lived things are expected.

However this logic doesn't apply to `&mut`. To see why `&mut` should be invariant over T, consider the following code:

    fn overwrite<T: Copy>(input: &mut T, new: &mut T) {
        *input = *new;
    }

    fn main() {
        let mut forever_str: &'static str = "hello";
        {
            let string = String::from("world");
            overwrite(&mut forever_str, &mut &*string);
        }
        // Oops, printing free'd memory
        println!("{}", forever_str);
    }

The signature of `overwrite` is clearly valid: it takes mutable references to two values of the same type, and overwrites one with the other.

But, if `&mut T` was covariant over T, then `&mut &'static str` would be a subtype of `&mut &'a str`, since `&'static str` is a subtype of `&'a str`. Therefore the lifetime of `forever_str` would successfully be "shrunk" down to the shorter lifetime of `string`, and `overwrite` would be called successfully. `string` would subsequently be dropped, and `forever_str` would point to freed memory when we print it! Therefore `&mut` should be invariant.

This is the general theme of variance vs invariance: if variance would allow you to store a short-lived value in a longer-lived slot, then invariance must be used.

More generally, the soundness of subtyping and variance is based on the idea that its ok to forget details, but with mutable references there's always someone (the original value being referenced) that remembers the forgotten details and will assume that those details haven't changed. If we do something to invalidate those details, the original location can behave unsoundly.

However it is sound for `&'a mut T` to be covariant over `'a`. The key difference between `'a` and T is that `'a` is a property of the reference itself, while T is something the reference is borrowing. If you change T's type, then the source still remembers the original type. However if you change the lifetime's type, no one but the reference knows this information, so it's fine. Put another way: `&'a mut T` owns `'a`, but only borrows T.

`Box` and `Vec` are interesting cases because they're covariant, but you can definitely store values in them! This is where Rust's typesystem allows it to be a bit more clever than others. To understand why it's sound for owning containers to be covariant over their contents, we must consider the two ways in which a mutation may occur: by-value or by-reference.

If mutation is by-value, then the old location that remembers extra details is moved out of, meaning it can't use the value anymore. So we simply don't need to worry about anyone remembering dangerous details. Put another way, applying subtyping when passing by-value destroys details forever. For example, this compiles and is fine:

    # #![allow(unused_variables)]
    #fn main() {
    fn get_box<'a>(str: &'a str) -> Box<&'a str> {
        // String literals are `&'static str`s, but it's fine for us to
        // "forget" this and let the caller think the string won't live that long.
        Box::new("hello")
    }
    #}

If mutation is by-reference, then our container is passed as `&mut Vec<T>`. But `&mut` is invariant over its value, so `&mut Vec<T>` is actually invariant over `T`. So the fact that `Vec<T>` is covariant over `T` doesn't matter at all when mutating by-reference.

But being covariant still allows `Box` and `Vec` to be weakened when shared immutably. So you can pass a `&Vec<&'static str>` where a `&Vec<&'a str>` is expected.

The invariance of the cell types can be seen as follows: `&` is like an `&mut` for a cell, because you can still store values in them through an `&`. Therefore cells must be invariant to avoid lifetime smuggling.

`fn` is the most subtle case because they have mixed variance, and in fact are the only source of contravariance. To see why `fn(T) -> U` should be contravariant over T, consider the following function signature:

    // 'a is derived from some parent scope
    fn foo(&'a str) -> usize;

This signature claims that it can handle any `&str` that lives at least as long as `'a`. Now if this signature was covariant over `&'a str`, that would mean

    fn foo(&'static str) -> usize;

could be provided in its place, as it would be a subtype. However this function has a stronger requirement: it says that it can only handle `&'static str`s, and nothing else. Giving `&'a str`s to it would be unsound, as it's free to assume that what it's given lives forever. Therefore functions definitely shouldn't be covariant over their arguments.

However if we flip it around and use contravariance, it does work! If something expects a function which can handle strings that live forever, it makes perfect sense to instead provide a function that can handle strings that live for less than forever. So

    fn foo(&'a str) -> usize;

can be passed where

    fn foo(&'static str) -> usize;

is expected.

To see why `fn(T) -> U` should be covariant over U, consider the following function signature:

    // 'a is derived from some parent scope
    fn foo(usize) -> &'a str;

This signature claims that it will return something that outlives `'a`. It is therefore completely reasonable to provide

    fn foo(usize) -> &'static str;

in its place, as it does indeed return things that outlive `'a`. Therefore functions are covariant over their return type.

`*const` has the exact same semantics as `&`, so variance follows. `*mut` on the other hand can dereference to an `&mut` whether shared or not, so it is marked as invariant just like cells.

This is all well and good for the types the standard library provides, but how is variance determined for type that you define? A struct, informally speaking, inherits the variance of its fields. If a struct `Foo` has a generic argument `A` that is used in a field `a`, then Foo's variance over `A` is exactly `a`'s variance. However if `A` is used in multiple fields:

- If all uses of A are covariant, then Foo is covariant over A
- If all uses of A are contravariant, then Foo is contravariant over A
- Otherwise, Foo is invariant over A

    # #![allow(unused_variables)]
    #fn main() {
    use std::cell::Cell;

    struct Foo<'a, 'b, A: 'a, B: 'b, C, D, E, F, G, H, In, Out, Mixed> {
        a: &'a A,     // covariant over 'a and A
        b: &'b mut B, // covariant over 'b and invariant over B

        c: *const C,  // covariant over C
        d: *mut D,    // invariant over D

        e: E,         // covariant over E
        f: Vec<F>,    // covariant over F
        g: Cell<G>,   // invariant over G

        h1: H,        // would also be variant over H except...
        h2: Cell<H>,  // invariant over H, because invariance wins all conflicts

        i: fn(In) -> Out,       // contravariant over In, covariant over Out

        k1: fn(Mixed) -> usize, // would be contravariant over Mixed except..
        k2: Mixed,              // invariant over Mixed, because invariance wins all conflicts
    }
    #}