Module qcell::doctest_tcell

This tests the TCell implementation.

It's not possible to have two simultaneous owners for the same marker type:

struct Marker;
type ACellOwner = TCellOwner<Marker>;
let mut owner1 = ACellOwner::new();
let mut owner2 = ACellOwner::new();  // Panics here

It should be impossible to copy a TCellOwner:

type ACell<T> = TCell<Marker, T>;
type ACellOwner = TCellOwner<Marker>;
let mut owner1 = ACellOwner::new();
let mut owner2 = owner1;
let rc = Rc::new(owner1.cell(100u32));  // Compile fail

It should be impossible to clone a TCellOwner:

let mut owner1 = ACellOwner::new();
let owner2 = owner1.clone();  // Compile fail

Two different owners can't borrow each other's cells immutably:

struct MarkerA;
type ACellOwner = TCellOwner<MarkerA>;
type ACell<T> = TCell<MarkerA, T>;
struct MarkerB;
type BCellOwner = TCellOwner<MarkerB>;
type BCell<T> = TCell<MarkerB, T>;

let mut owner_a = ACellOwner::new();
let mut owner_b = BCellOwner::new();
let c1 = Rc::new(ACell::new(100u32));

let c1ref = owner_b.ro(&*c1);   // Compile error
println!("{}", *c1ref);

Or mutably:

let mut owner_a = ACellOwner::new();
let mut owner_b = BCellOwner::new();
let c1 = Rc::new(ACell::new(100u32));

let c1mutref = owner_b.rw(&*c1);    // Compile error
println!("{}", *c1mutref);

You can't have two separate mutable borrows active on the same owner at the same time:

let mut owner = ACellOwner::new();
let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

let c1mutref = owner.rw(&c1);
let c2mutref = owner.rw(&c2);  // Compile error
*c1mutref += 1;
*c2mutref += 2;

However with rw2() you can do two mutable borrows at the same time, since this call checks at runtime that the two references don't refer to the same memory:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

let (c1mutref, c2mutref) = owner.rw2(&c1, &c2);
*c1mutref += 1;
*c2mutref += 2;
assert_eq!(303, owner.ro(&c1) + owner.ro(&c2));   // Success!

You can't have a mutable borrow at the same time as an immutable borrow:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

let c1ref = owner.ro(&c1);
let c1mutref = owner.rw(&c1);    // Compile error
println!("{}", *c1ref);

Not even if it's borrowing a different object:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

let c1mutref = owner.rw(&c1);
let c2ref = owner.ro(&c2);    // Compile error
*c1mutref += 1;

Many immutable borrows at the same time is fine:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

let c1ref = owner.ro(&c1);
let c2ref = owner.ro(&c2);
let c1ref2 = owner.ro(&c1);
let c2ref2 = owner.ro(&c2);
assert_eq!(600, *c1ref + *c2ref + *c1ref2 + *c2ref2);   // Success!

Whilst a reference is active, it's impossible to drop the Rc:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

let c1ref = owner.ro(&c1);
drop(c1);    // Compile error
println!("{}", *c1ref);

Also, whilst a reference is active, it's impossible to call anything else that uses the owner in an incompatible way, e.g. &mut when there's a & reference:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

fn test(o: &mut ACellOwner) {}

let c1ref = owner.ro(&c1);
test(&mut owner);    // Compile error
println!("{}", *c1ref);

Or & when there's a &mut reference:

let c1 = Rc::new(ACell::new(100u32));
let c2 = Rc::new(ACell::new(200u32));

fn test(o: &ACellOwner) {}

let c1mutref = owner.rw(&c1);
test(&owner);    // Compile error
*c1mutref += 1;

TCellOwner and TCell should be both Send and Sync by default:

struct Marker;
fn is_send_sync<T: Send + Sync>() {}
is_send_sync::<TCellOwner<Marker>>();
is_send_sync::<TCell<Marker, ()>>();

So for example we can share a cell ref between threads (Sync), and pass an owner back and forth (Send):

type ACellOwner = TCellOwner<Marker>;
type ACell = TCell<Marker, i32>;

let mut owner = ACellOwner::new();
let cell = ACell::new(100);

*owner.rw(&cell) += 1;
let cell_ref = &cell;
let mut owner = crossbeam::scope(move |s| {
    s.spawn(move |_| {
        *owner.rw(cell_ref) += 2;
        owner
    }).join().unwrap()
}).unwrap();
*owner.rw(&cell) += 4;
assert_eq!(*owner.ro(&cell), 107);

However you can't send a cell that's still borrowed:

let owner = ACellOwner::new();
let cell = ACell::new(100);
let val_ref = owner.ro(&cell);
std::thread::spawn(move || {
    assert_eq!(*owner.ro(&cell), 100);
}).join();
assert_eq!(*val_ref, 100);

If the contained type isn't Sync, though, then TCell shouldn't be Sync either:

fn is_sync<T: Sync>() {}
is_sync::<TCell<Marker, Cell<i32>>>();  // Compile fail

type ACellOwner = TCellOwner<Marker>;
type ACell = TCell<Marker, Cell<i32>>;

let owner = ACellOwner::new();
let cell = ACell::new(Cell::new(100));

// This would be a data race if the compiler permitted it, but it doesn't
std::thread::spawn(|| owner.ro(&cell).set(200));  // Compile fail
owner.ro(&cell).set(300);

If the contained type isn't Send, the TCell should be neither Sync nor Send:

fn is_sync<T: Sync>() {}
is_sync::<TCell<Marker, Rc<()>>>();  // Compile fail

fn is_send<T: Send>() {}
is_send::<TCell<Marker, Rc<()>>>();  // Compile fail

type ACellOwner = TCellOwner<Marker>;
type ACell = TCell<Marker, Rc<i32>>;

let owner = ACellOwner::new();
let cell = ACell::new(Rc::new(100));

// We aren't permitted to move the Rc to another thread
std::thread::spawn(move || {    // Compile fail
    assert_eq!(100, **owner.ro(&cell));
}).join();