Getting "temporary value dropped while borrowed" when trying to update an Option<&str> in a loop

In your code, it remains unclear who the owner of the String referenced in last: Option<&str> is supposed to be. You could introduce an extra mutable local variable that owns the string. But then you would have two variables: the owner and the reference, which seems redundant. It would be much simpler to just make last the owner:

struct MyRes {
    str: String,
}

fn main() {
    let times = 10;
    let mut last: Option<String> = None;

    for _i in 0..times {
        last = do_something(&last).map(|r| r.str);
    }
}

fn do_something(_o: &Option<String>) -> Option<MyRes> {
    Some(MyRes {
        str: "whatever string".to_string(),
    })
}

In do_something, you can just pass the whole argument by reference, this seems more likely to be what you wanted. Also note that naming your own struct Result is a bad idea, because Result is such a pervasive trait built deeply into the compiler (?-operator etc).

---

Follow-up question: Option<&str> or Option<String>?

Both Option<&str> and Option<String> have different trade-offs. One is better for passing string literals, other is better for passing owned Strings. I'd actually propose to use neither, and instead make the function generic over type S that implements AsRef<str>. Here is a comparison of various methods:

fn do_something(o: &Option<String>) {
    let _a: Option<&str> = o.as_ref().map(|r| &**r);
    let _b: Option<String> = o.clone();
}
fn do_something2(o: &Option<&str>) {
    let _a: Option<&str> = o.clone(); // do you need it?
    let _b: Option<String> = o.map(|r| r.to_string());
}
fn do_something3<S: AsRef<str>>(o: &Option<S>) {
    let _a: Option<&str> = o.as_ref().map(|s| s.as_ref());
    let _b: Option<String> = o.as_ref().map(|r| r.as_ref().to_string());
}

fn main() {
    let x: Option<String> = None;
    let y: Option<&str> = None;

    do_something(&x);                           // nice
    do_something(&y.map(|r| r.to_string()));    // awkward & expensive

    do_something2(&x.as_ref().map(|x| &**x));   // cheap but awkward
    do_something2(&y);                          // nice

    do_something3(&x);                          // nice
    do_something3(&y);                          // nice, in both cases
}

Note that not all of the above combinations are very idiomatic, some are added just for completeness (e.g. asking for AsRef<str> and then building an owned String out of seems a bit strange).

r.str.to_owned() is a temporary value. You can take a reference to a temporary, but because the temporary value will usually be dropped (destroyed) at the end of the innermost enclosing statement, the reference becomes dangling at that point. In this case the "innermost enclosing statement" is either the last line of the loop, or the loop body itself -- I'm not sure exactly which one applies here, but it doesn't matter, because either way, you're trying to make last contain a reference to a String that will soon be dropped, making last unusable. The compiler is right to stop you from using it again in the next iteration of the loop.

The easiest fix is just to not make last a reference at all -- in the example, it's not necessary or desirable. Just use Option<String>:

fn main() {
    let times = 10;
    let mut last = None;

    for _ in 0..times {
        last = match do_something(last) {
            Some(r) => Some(r.str),
            None => None,
        };
    }
}

fn do_something(_: Option<String>) -> Option<Result> {
    // ...
}

There are also ways to make the reference version work; here is one:

let mut current;  // lift this declaration out of the loop so `current` will have
                  // a lifetime longer than one iteration
for _ in 0..times {
    current = do_something(last);
    last = match current {
        Some(ref r) => Some(&r.str),  // borrow from `current` in the loop instead
                                      // of from a newly created String
        None => None,
    };
}

You might want to do this if your code is more complicated than the example and using String would mean a lot of potentially expensive .clone()s.