Is something like possible in rust?
fn replace_abracadabra(s: &str) -> &str {
if s.starts_with("abracadabra") {
"i am static" + s.chars().last().unwrap()
} else {
s
}
}
Here, "i am static" + s.chars().last().unwrap() is all borrowed, but is now a non-contiguous block of memory.
>Solution :
No, slices (both string and array slices) are contiguous by definition.
You also cannot borrow unless you have an owned value to borrow from. In this case you are trying to construct a new value and then return a reference to it, which is not possible in Rust.
The two pieces of information you’re trying to concatenate are indeed borrowed, but the concatenation creates a new value that contains non-borrowed information. Even if you let the string data live where it is, the information "glue these two strings together" cannot be borrowed. You could return a (&str, &str) for example — the tuple is an owned value that contains two borrowed slices in it.
This function should probably return an owned String value. You could also return a Cow<'_, str> which would allow you to borrow s in the else block without copying, but return an owned value in the first case.