Why don't I have to declare that x is reusable/duplicable with affine semantics and function types?

Question

I was told that Rust has a semantics in affine logic -- so one has deletion/weakening but not duplication/contraction.

The following compiles:

fn throw_away<A, B>(x: A, _y: B) -> A {
    x
}

Because duplication is disallowed, the following does not compile:

fn dup<A>(x: A) -> (A, A) {
    (x, x)
}

Similarly, neither of these compile:

fn throw_away3<A, B>(x: A, f: fn(A) -> B) -> A {
    x;
    f(x)
}

fn throw_away4<A, B>(x: A, f: fn(A) -> B) -> A {
    throw_away(x, f(x))
}

Weakening is also witnessable

fn weaken<A, B, C>(f: fn(A) -> B) -> impl Fn(A, C) -> B {
    move |x: A, y: C| f(x)
}

Instead of returning fn(A, C) -> B, we returned impl Fn(A, C) -> B. Is there a way to return fn(A, C) -> B instead? It's fine if not; I'm just curious.

Something else I expect is that you can lift A to () -> A. However, functions in Rust can be duplicated and used more than once. For example,

fn app_twice(f: fn(A) -> A, x: A) -> A {
    f(f(x))
}

Suppose there was actually a function lift(x: A) -> fn() -> A, then we could break the move semantics. For example, this would allow

fn dup_allowed(x: A) -> (A, A) {
    let h = lift(x);
    (h(), h())
}

Thus to lift A to fn() -> A, we need to know that the function is "linear/affine" or can be used only once. Rust provides a type for this: FnOnce() -> A. In the following, the first compiles, and the second does not.

fn app_once(f: impl FnOnce(A) -> A, x: A) -> A {
    f(x)
}

fn app_twice2(f: impl FnOnce(A) -> A, x: A) -> A {
    f(f(x))
}

The following functions are inverses of each other (probably, I don't know Rust's semantics well enough to say that they are actually inverse to each other):

fn lift_up<A>(x: A) -> impl FnOnce() -> A {
    move || x
}

fn lift_up_r<A>(f: impl FnOnce() -> A) -> A {
    f()
}

Since fn dup<A>(x: A) -> (A, A) { (x,x) } does not compile, I thought that the following might be a problem:

fn dup<A>(x: fn() -> A) -> (A, A) {
    (x(), x())
}

It seems that Rust is doing something special for fn(A) -> B types.

Why don't I have to declare that x is reusable/duplicable in the above?

Perhaps something different is going on. Declared functions are a bit special fn f(x: A) -> B { ... } is a particular witness that A -> B. Thus if f needs to be used multiple times, it can be reproved as many times as needed, but fn(A) -> B is a completely different thing: it is not a constructed thing but a hypothetical thing, and must be using that fn(A) -> Bs are duplicatable. In fact, I've been thinking that it's more like a freely duplicable entity. Here's my rough analogy:

• fn my_fun<A,B>(x :A) -> B { M } "is" x:A |- M:B
• fn(A) -> B "is" !(A -o B) hence freely duplicable
• Thus fn() -> A "is" !(() -o A) = !A hence fn () -> A is the (co)free duplication on A
• fn dup_arg<A: Copy>(x: A) -> B { M } "says" that A has duplication or is a comonoid
• impl FnOnce (A) -> B "is" A -o B

But this can't be right... For what is impl Fn(A) -> B? From playing around a bit, it seems that fn(A) -> B is more strict than Fn(A) -> B. What am I missing?

Aileneaileron

You can duplicate fns at will because they are Copy, meaning they hold only trivially cloneable information, just like, say, numbers. A fn is just the address of the function in the executable, and copying it only entail duplicating the pointer. This is not the case with closures, which may capture arbitrary state, as your lift example does. – Gillispie

Thanks! When I declare a function at the top level like fn my_fun(x:A) ... then is rust treating references to my_fun as addresses/pointers in the same way that it's treating fn(A) -> B passed in as an argument to a function? – Aileneaileron

fn(...) are always simple references to global data. Types that implement the Fn(...) trait, though, don't need to be. If you pass an fn() to generic code (that can work with any Fn, not just a concrete fn), then inside generic code you can't duplicate them unless you explicitly specify + Copy (in which case the function returned by lift won't qualify). impl Fn(...) is an anonymous type only known to implement the Fn(...) trait. Your question was actually a very interesting read, but I'm not sure what the actual question is. – Gillispie

Answer 1

fn weaken<A, B, C>(f: fn(A) -> B) -> impl Fn(A, C) -> B {
    move |x: A, y: C| f(x)
}

Instead of returning fn(A, C) -> B, we returned impl Fn(A, C) -> B. Is there a way to return fn(A, C) -> B instead? It's fine if not; I'm just curious.

No, because a fn is by definition not a closure: it cannot contain any state that wasn't compiled into the program (in this case, the value of f). This is closely related to your next observation: because a fn cannot close over anything, it trivially cannot contain any non-Copy types and therefore can always be called multiple times, or itself copied, without violating the properties we're discussing.

Precisely: all fn(..) -> _ types implement Fn and Copy (as well as FnOnce).

• Copy is the marker trait ('marker' meaning it provides no methods) that has the special purpose of telling the compiler that it is free to copy the bits of a type automatically whenever it is used more than once. Anything implementing Copy is opting out of the move-but-not-copy system -- but can't thereby violate the non-Copy-ness of a different type.
• Fn is the trait for functions that can be called by immutable reference (not modifying or consuming the function itself). This is in principle separate from Copy, but it's very similar in effect; the differences that one could end up with (some of these can't happen in ordinary code) are:
  • If a function implements Fn but not Copy or Clone, then you can't store the function multiple places but you can call it as many times as you want.
  • If a function implements Copy but not Fn (only FnOnce), then this is invisible because every call of it (except for the last) implicitly copies it.
  • If a function implements Clone but not Fn or Copy, then you would have to .clone() it each time you called it (except the last).

---

And indeed the following functions are inverses of eachother (probably, I don't know rust's semantics well enough to say that they are actually inverse to each other):

fn lift_up<A> (x:A) -> impl FnOnce () -> A {move | | x}
fn lift_up_r<A> (f : impl FnOnce () -> A) -> A {f()}

lift_up_r accepts functions that lift_up did not produce; for example, if f has a side effect, panics, or hangs then let f = lift_up(lift_up_r(f)); has that effect. Ignoring that, they are inverses. An even better pair of inverses without that caveat would be functions for moving a value into a struct and back out -- which this is effectively doing, except for allowing inputs that aren't of that particular struct type.

---

Since fn dup (x:A) -> (A,A) {(x,x)} does not compile, I thought that the following might be a problem:

fn dup<A> (x : fn() -> A) -> (A,A) {(x(),x()}

But it seems that rust is doing something special for fn(A) -> B types. Finally, my question: why don't I have to declare that x is reusable/duplicable in the above?

When you have a generic function with a type variable, fn dup<A>, the compiler makes no assumptions about the properties of A (except that it is Sized unless you opt out of that implicit bound, because working with non-Sized values is highly restrictive and usually not what you want). In particular, it does not assume that A implements Copy.

On the other hand, as I mentioned above, all fn types implement Fn and Copy, so they can always be duplicated and reused.

The way to write a dup function which operates on general functions and fails to compile in the way you expect is:

fn dup<A, F>(x: F) -> (A,A)
where
    F: FnOnce() -> A
{
    (x(),x())
}

Here, we tell the compiler that F is a type of function which is consumed by calling it, and don't tell it about any way to duplicate F. So, it fails to compile with "error[E0382]: use of moved value: x". The shortest way to make this compile would be to add the bound F: Copy, and the most general would be to add F: Clone and an explicit .clone() call.

---

Perhaps something different is going on. Declared functions are a bit special fn f(x:A) -> B {...} is a particular witness that A -> B. Thus if f needs to be used multiple times, it can be reproved as many times as needed. But fn(A) -> B is a completely different thing: it is not a constructed thing but a hypothetical thing, and must be using a that fn(A) -> Bs are duplicatable. In fact, I've been thinking that it's more like a freely duplicable entity.

I'm no logician, but I think that the first half of this is not correct. In particular, (outside of some irrelevant considerations about generics) there are no properties that "a declared function" has that an arbitrary value of type fn(A) -> B does not have. Rather, the value of type fn(A) -> B can be copied, and that copiability corresponds directly to the fact that "it can be reproved", because (until we start introducing ideas like JIT code generation) every value of type fn(A) -> B refers to a compiled piece of code (and no other data) -- and hence a lemma that the compiler has checked and given the program license to reuse it as many times as needed at run time.

For what is impl Fn(A) -> B? From playing around a bit, it seems that fn(A) -> B is more strict than Fn(A) -> B. What am I missing?

The impl syntax serves different roles, but in argument position it is almost exactly a shorthand for generics. If I write

fn foo<A, B>(f: impl Fn(A) -> B) {}

then that is equivalent to

fn foo<A, B, F>(f: F) 
where
   F: Fn(A) -> B
{}

except that the caller is not allowed to specify any of the parameters when any impl argument types exist (this is not relevant to your interests but I mention it for accuracy). Thus, we're telling the compiler that F can be anything as long as it is callable as a reusable function. In particular, we're not specifying F: Copy or F: Clone. fn(A) -> B, on the other hand, is a concrete type which implements Fn(A) -> B and Copy, so you get that for free.

In return position, fn ... -> impl Fn(A) -> B, the impl denotes an existential type: you're asserting that there exists some type implementing Fn which the function will return. The compiler tracks the concrete type in order to generate code, but your program avoids naming it. This is necessary when returning a closure, but optional when returning a function that does not close over anything: for example, you can write

fn foo<A>() -> fn(A) -> A {
    |x| x
}

Spermato

Wow, thanks for the amazing and detailed reply! One thing I'm confused by is that every fn(A) -> B is a compile time constant. This is a really neat property! Is this because fn(A) -> B is just a pointer and at compile time is given a constant address? Suppose we we write a function "fn w<A> (x:i32) -> fn(A) -> A {...}" where if x <0 it's "|x| x" and otherwise is something like "|x| while(true){}x". Then in some other function generic over A we could have some variable "y:i32" and then "let h : fn(A) -> A = w(y); ...". In that case is h deduced to be a constant? – Aileneaileron

I'm going to wait till tomorrow to accept your answer so that I can set a bounty and award it. I'm grateful for your answer and your patience with my novice and boring question. – Aileneaileron

@JonathanGallagher "Is this because fn(A) -> B is just a pointer and at compile time is given a constant address?" — The address is not strictly a compile time constant because it may be modified (in fact, is likely randomized) by the operating system's program loader/dynamic linker — but it won't change over the lifetime of the process. But those are all more or less implementation details; the important thing is that the definition of the fn type includes that it implements Copy, and the system does what is necessary to ensure that. – Spermato

wait till tomorrow to accept your answer so that I can set a bounty and award it @JonathanGallagher you can award a bounty even if a question has been answered / accepted. – Venita