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Last active January 12, 2022 00:14
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// Borrowing 101. Easy to see that the borrows don't overlap
// because everything is explicit and has a mapping to a
// lexical range in the source.
#[derive(Debug)]
struct NotCopy;
fn main() {
// owner (can move the value if it wants, responsible for calling destructor)
let mut obj = NotCopy {};
let immutable_borrow = &obj;
let mutable_borrow = &mut obj;
dbg!(mutable_borrow);
// Uncomment this to get a borrow checker error.
// dbg!(immutable_borrow);
}
// Borrowing 501. How long a borrow lasts is not strictly a lexical
// construct. Note that Rust references behave more similar to
// C pointers than C++ references.
#[derive(Debug)]
struct NotCopy<'a> {
num: i8,
/// References are always valid. For self reference we need
/// an initial state that doesn't have the reference so we
/// wrap it in Option.
///
/// The borrow lasts for an arbitrary lifetime of 'a. We
/// are at the declaration site so we don't have information
/// as to exactly how long that is, merely a variable for now.
borrow_slot: Option<&'a i8>,
}
fn main() {
// owner (can move the value if it wants, responsible for calling destructor)
let obj = NotCopy {
num: 0,
borrow_slot: None,
};
// move it, just for fun.
let mut same_obj = obj;
// Can't use obj anymore, cause it moved to same_obj.
// The following causes a compile error
// dbg!(obj);
// Make the struct borrow from itself (!!)
// This fills in the 'a lifetime variable declared in the struct. How long does
// the borrow lasts? As long as the lifetime of the ownership in this case
// because the struct borrows from itself.
same_obj.borrow_slot = Some(&same_obj.num);
// The point: unlike in lexical.rs, the immutable borrow (the `&`) from line above
// lasts for _longer_ than the line it is on.
//
// Because there is still a borrow outstanding, we can't move same_obj
// anymore.
// let _illegal_move = same_obj;
// We can't move it, but we can still borrow it.
dbg!(&same_obj);
}
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