Two lifetime clarification

In the first change "this" and "that" is slightly ambiguous, at least for a
first introduction. Actually, I'm not 100% sure I got it right, but at least, if
I got it wrong, it proves my point it needed clarification

The second change is simply because "x has to live (for 'b to be printed)"
seemeds to mean that "'b" will be printed. I hope it makes clear that the actual
parsing is "(x has to live for 'b) in order to be printed"
pull/350/head
Arthur Milchior 3 years ago
parent c7d8467ca9
commit 07d6fcb294

@ -80,8 +80,8 @@ z = y;
'b: {
let z: &'b i32;
'c: {
// Must use 'b here because this reference is
// being passed to that scope.
// Must use 'b here because the reference to x is
// being passed to the scope 'b.
let y: &'b i32 = &'b x;
z = y;
}
@ -208,11 +208,12 @@ violate the *second* rule of references.
However this is *not at all* how Rust reasons that this program is bad. Rust
doesn't understand that `x` is a reference to a subpath of `data`. It doesn't
understand `Vec` at all. What it *does* see is that `x` has to live for `'b` to
be printed. The signature of `Index::index` subsequently demands that the
reference we take to `data` has to survive for `'b`. When we try to call `push`,
it then sees us try to make an `&'c mut data`. Rust knows that `'c` is contained
within `'b`, and rejects our program because the `&'b data` must still be alive!
understand `Vec` at all. What it *does* see is that `x` has to live for `'b` in
order to be printed. The signature of `Index::index` subsequently demands that
the reference we take to `data` has to survive for `'b`. When we try to call
`push`, it then sees us try to make an `&'c mut data`. Rust knows that `'c` is
contained within `'b`, and rejects our program because the `&'b data` must still
be alive!
Here we see that the lifetime system is much more coarse than the reference
semantics we're actually interested in preserving. For the most part, *that's

Loading…
Cancel
Save