How to return a &str, avoiding "temporary value dropped while borrowed"

Question

I'm having trouble unwrapping Options and converting the values to string slices.

When the Option has Some value, I want to convert it to a &str.
When the Option is None, I simply return an empty string slice "".

For example:

let val: Option<u8> = row.get(k);
match val {
    None => "",
    Some(v) => v.to_string().as_str()
}

But this fails with errors like this.

temporary value dropped while borrowed
consider using a `let` binding to create a longer lived value rustc E0716
main.rs(.....): temporary value is freed at the end of this statement

I have tried breaking it down in various different ways, for example like this

let val: Option<u8> = row.get(k);
match val {
    None => "",
    Some(v) => {
        let vstr = v.to_string();
        vstr.as_str()
    }
}

But I'm still really stuck. I've looked through a many similar problems but none are quite the same, and I'm still stuck. How can I do this?

Clarification, I want the decimal representation of the tinyint, not the character point. This is not utf8 string, just a number.

let x: u8 = 10;
let s: String = x.to_string();
println!("10 = {}", s);

This should print 10 = 10

Or, is it really `Option` and maybe you want to convert that 8 bit integer into it's decimal representation in a string? — harmic, Apr 26 '21 at 01:30
@antoninGRAVIL, no not really meant to be a byte char. think of it as a tinyint, harmic, yes more like a decimal representation in a string is what I want. — Christopher Plewright, Apr 26 '21 at 01:41
it's because you are borrowing from `vstr` which is dropped when it goes out of scope. Since you are returning a reference, that reference must point to something that outlives the reference. Try returning a `String` instead of an `&str`. — Mel, Apr 26 '21 at 01:55
@Mel YES! by using String I'm able to work around it for now. — Christopher Plewright, Apr 26 '21 at 02:15
Does this answer your question? [Proper way to return a new string in Rust](https://stackoverflow.com/questions/43079077/proper-way-to-return-a-new-string-in-rust) — Jmb, Apr 26 '21 at 06:32
@Jmb, I'd say that link only indirectly helps, because it adds to my understanding of the underlying concepts, ie; lifetime and ownership. I'm not returning a string. Thanks for sharing it. — Christopher Plewright, Apr 27 '21 at 08:50

score 3 · Accepted Answer · answered Apr 26 '21 at 02:20

If you're trying to use the &str in the same function, you can hoist the declaration so that the string lives long enough:

let val: Option<u8> = row.get();
let vstr;
let msg = match val {
    None => "",
    Some(v) => {
        vstr = v.to_string();
        vstr.as_str()
    }
};
println!("value is {}", msg);

If your goal is to return the &str to a caller, then that's not possible. That's a borrowed value, and the string that owns the memory won't live past the end of the function at most. You can either return String or Cow<'static, str>.

For example, you could do this:

let val: Option<u8> = row.get();
match val {
    None => "".to_owned(),
    Some(v) => v.to_string(),
}

or this:

let val: Option<u8> = row.get();
match val {
    None => "".into(),
    Some(v) => v.to_string().into(),
}

The last snippet seems to be missing type ascription, e.g. `let msg: Cow<'static, str> = match val { ... }`. — user4815162342, Apr 26 '21 at 20:46

score 0 · Answer 2 · answered Apr 26 '21 at 07:04

In the more general case, when you cannot keep the string around, you could use Cow:

let val: Option<u8> = row.get(k);
let msg = match val {
    None => Cow::Borrowed(""),
    Some(v) => Cow::Owned(v.to_string()),
};

Now msg holds either a borrowed or an owned string, depending on the situation.

How to return a &str, avoiding "temporary value dropped while borrowed"

2 Answers2