[−][src]Module qcell::doctest_lcell
This tests the LCell
implementation.
It should be impossible to copy a &mut LCellOwner
:
LCellOwner::scope(|mut owner1| { let owner2 = owner1; let rc = Rc::new(owner1.cell(100u32)); // Compile fail });
It should be impossible to clone an LCellOwner:
LCellOwner::scope(|mut owner1| { let owner2 = owner1.clone(); // Compile fail });
Two different owners can't borrow each other's cells immutably:
LCellOwner::scope(|mut owner1| { LCellOwner::scope(|mut owner2| { let c1 = Rc::new(LCell::new(100u32)); let c1ref1 = owner1.ro(&c1); let c1ref2 = owner2.ro(&c1); // Compile error println!("{}", *c1ref2); }); });
Or mutably:
LCellOwner::scope(|mut owner1| { LCellOwner::scope(|mut owner2| { let c1 = Rc::new(owner1.cell(100u32)); let c1mutref2 = owner2.rw(&c1); // Compile error println!("{}", *c1mutref2); }); });
You can't have two separate mutable borrows active on the same owner at the same time:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); let c2 = Rc::new(LCell::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:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); let c2 = Rc::new(LCell::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:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); let c2 = Rc::new(LCell::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:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); let c2 = Rc::new(LCell::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:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); let c2 = Rc::new(LCell::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
:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); 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:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); fn test(o: &mut LCellOwner) {} let c1ref = owner.ro(&c1); test(&mut owner); // Compile error println!("{}", *c1ref); });
Or &
when there's a &mut
reference:
LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); fn test(o: &LCellOwner) {} let c1mutref = owner.rw(&c1); test(&owner); // Compile error *c1mutref += 1; });
Two examples of passing owners and cells in function arguments. This needs lifetime annotations.
use qcell::{LCell, LCellOwner}; use std::rc::Rc; LCellOwner::scope(|mut owner| { let c1 = Rc::new(LCell::new(100u32)); fn test<'id>(o: &mut LCellOwner<'id>, rc: &Rc<LCell<'id, u32>>) { *o.rw(rc) += 1; } test(&mut owner, &c1); let c1mutref = owner.rw(&c1); *c1mutref += 1; });
use qcell::{LCell, LCellOwner}; use std::rc::Rc; LCellOwner::scope(|mut owner| { struct Context<'id> { owner: LCellOwner<'id>, } let c1 = Rc::new(LCell::new(100u32)); let mut ct = Context { owner }; fn test<'id>(ct: &mut Context<'id>, rc: &Rc<LCell<'id, u32>>) { *ct.owner.rw(rc) += 1; } test(&mut ct, &c1); let c1mutref = ct.owner.rw(&c1); *c1mutref += 2; });
LCellOwner
and LCell
should be both Send
and Sync
by default:
fn is_send_sync<T: Send + Sync>() {} is_send_sync::<LCellOwner<'_>>(); is_send_sync::<LCell<'_, ()>>();
So for example we can share a cell ref between threads (Sync), and pass an owner back and forth (Send):
LCellOwner::scope(|mut owner| { let cell = LCell::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:
LCellOwner::scope(|mut owner| { let cell = LCell::new(100); let val_ref = owner.ro(&cell); crossbeam::scope(move |s| { s.spawn(move |_| assert_eq!(*owner.ro(&cell), 100)).join().unwrap(); // Compile fail }).unwrap(); assert_eq!(*val_ref, 100); });
If the contained type isn't Sync
, though, then LCell
shouldn't
be Sync
either:
fn is_sync<T: Sync>() {} is_sync::<LCell<'_, Cell<i32>>>(); // Compile fail
LCellOwner::scope(|owner| { let cell = LCell::new(Cell::new(100)); // This would likely be a data race if it compiled crossbeam::scope(|s| { // Compile fail let handle = s.spawn(|_| owner.ro(&cell).set(200)); owner.ro(&cell).set(300); handle.join().unwrap(); }).unwrap(); });
If the contained type isn't Send
, the LCell
should be neither
Sync
nor Send
:
fn is_sync<T: Sync>() {} is_sync::<LCell<'_, Rc<()>>>(); // Compile fail
fn is_send<T: Send>() {} is_send::<LCell<'_, Rc<()>>>(); // Compile fail
LCellOwner::scope(|owner| { let cell = LCell::new(Rc::new(100)); // We aren't permitted to move the Rc to another thread crossbeam::scope(move |s| { s.spawn(move |_| assert_eq!(100, **owner.ro(&cell))).join().unwrap(); // Compile fail }).unwrap(); });