Then what does for<‘a> mean? ELI5
Conversation
It's like in math/logic where `x` means "any, but a fixed x" where `forall` means "any possible lifetime you could come up with".
1
1
Because you instruct the lifetime checker that it must prove all possible configurations safe. It's used especially for closures where the caller is not yet known.
1
2
If you use this feature wrong the right way, you will actually get error messages that give counter examples through inductive proof. "Assume a lifetime '0 that is allowed, there's a lifetime '1..."
1
2
It may be more accessible this way: If you have a function `fn x<'a>(&'a Something)` and you call it, the compiler has a concrete lifetime (the one of Something) to check. So it fills in the lifetime parameter through the actual time that Something lives.
1
But in the case of closures and some other things (which may be defined, but called _later_ in a function-like manner), we don't know the lifetime in the scope of the caller yet. So `for<'a>` means "independent of the situation at the call site, this must _always be safe_".
1
1
1
for<'a> is like a lambda for lifetimes; one will be provided later, and the type must work for any lifetime provided
3
1
Hmm just gotta be careful here, forall is not a lambda - it’s the type of a lambda! It gets a bit weird in Rust as they are only used within trait constraints, and you never see the corresponding lifetime functions, nor can you construct them yourself.
1
1
In hypothetical Rust:
let id : for<T> fn(x: T) -> T = <T>|x: T| -> T { x };
assert!(id::<i32>(3), 3);
assert!(id::<&str>("hi"), "hi");
(was meant to be `assert_eq!`)
See how `for` lets you write a `fn` type that depends on the type parameter `T` that is applied. Note that this is actually very similar to the `fn` type, just at the type level, and implicit.
2
This would be problematic for monomorphisation however, which is why you can't do it in Rust right now. I.e. depending on the type you apply at runtime you'd need to generate different code for `id`, which is 'difficult'. 😅
1
Show replies