[−][src]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();