What is the difference between '&self' and '&'a self'?
I recently had an error which was simply resolved by changing
impl<'a> Foo<'a> {
fn foo(&'a self, path: &str) -> Boo<'a> { /* */ }
}
to
impl<'a> Foo<'a> {
fn foo(&self, path: &str) -> Boo { /* */ }
}
which did not make sense according to my understanding, as I thought that the second version is exactly the same as the first with applied lifetime elision.
In case we introduce a new lifetime for the method this seems to be the case according this example from the nomicon.
fn get_mut(&mut self) -> &mut T; // elided
fn get_mut<'a>(&'a mut self) -> &'a mut T; // expanded
So what are the differences between this and my first code snipped.
Solution 1:
Lifetime 'a
in fn foo(&'a self, ...) ...
is defined for impl<'a>
, that is it is the same for all foo
calls.
Lifetime 'a
in fn get_mut<'a>(&'a mut self) ...
is defined for the function. Different calls of get_mut
can have different values for 'a
.
Your code
impl<'a> Foo<'a> { fn foo(&'a self, path: &str) -> Boo<'a> { /* */ } }
is not the expansion of elided lifetime. This code ties lifetime of borrow &'a self
to the lifetime of structure Foo<'a>
. If Foo<'a>
is invariant over 'a
, then self
should remain borrowed as long as 'a
.
Correct expansion of elided lifetime is
impl<'a> Foo<'a> {
fn foo<'b>(&'b self, path: &str) -> Boo<'b> { /* */ }
}
This code doesn't depend on variance of structure Foo
to be able to borrow self
for shorter lifetimes.
Example of differences between variant and invariant structures.
use std::cell::Cell;
struct Variant<'a>(&'a u32);
struct Invariant<'a>(Cell<&'a u32>);
impl<'a> Variant<'a> {
fn foo(&'a self) -> &'a u32 {
self.0
}
}
impl<'a> Invariant<'a> {
fn foo(&'a self) -> &'a u32 {
self.0.get()
}
}
fn main() {
let val = 0;
let mut variant = Variant(&val);// variant: Variant<'long>
let mut invariant = Invariant(Cell::new(&val));// invariant: Invariant<'long>
{
let r = variant.foo();
// Pseudocode to explain what happens here
// let r: &'short u32 = Variant::<'short>::foo(&'short variant);
// Borrow of `variant` ends here, as it was borrowed for `'short` lifetime
// Compiler can do this conversion, because `Variant<'long>` is
// subtype of Variant<'short> and `&T` is variant over `T`
// thus `variant` of type `Variant<'long>` can be passed into the function
// Variant::<'short>::foo(&'short Variant<'short>)
}
// variant is not borrowed here
variant = Variant(&val);
{
let r = invariant.foo();
// compiler can't shorten lifetime of `Invariant`
// thus `invariant` is borrowed for `'long` lifetime
}
// Error. invariant is still borrowed here
//invariant = Invariant(Cell::new(&val));
}
Playground link