What is the problem with the following code?

fn longest(x: &str, y: &str) -> &str {
    let result;
    if x.len() > y.len() {
        result = x;
    } else {
        result = y;
    }
    result
}

The problem with the provided Rust code is related to lifetimes of the borrowed references. The function longest is trying to return a reference to a string (&str). However, Rust's borrow checker does not know how long the references x or y will live for; hence, it cannot ascertain that the returned reference will be valid. To solve this, you need to assure Rust that the returned reference will live at least as long as the shortest of x and y lifetimes. This could be guaranteed by using explicit lifetime parameters. Here's the corrected solution:

fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
    if x.len() > y.len() {
        x
    } else {
        y
    }
}

Now Rust knows that all the references in this function's signature must have the same lifetime 'a, ensuring that the output will always be valid as long as the inputs stay valid.

Can x and y have different lifetimes?

Yes, in Rust, x and y can have different lifetimes. The lifetime of x and y depends on where they are declared and how long they are borrowed for. In the context of functions, you can specify different lifetimes for different parameters. This is useful when a function needs to convey that it accepts arguments with different lifetimes. Here is an example:

fn longest<'a, 'b>(x: &'a str, y: &'b str) -> &'a str {
    x
}

In the function above, the x and y parameters have different lifetimes 'a and 'b, respectively. The function returns a string slice with the same lifetime as x, neglecting the lifetime of y.

What if I could return either x or y? What’s the return value’s lifetime then?

If a function could return either x or y, then the lifetime of the return value would have to be as long as the shortest of the lifetimes of x or y. This is because the function cannot know ahead of time which one of x and y it will return, so it must assume that it could return either and hence be prepared for the one with the shorter lifetime. In Rust, you can express this by using a single lifetime parameter on all the references in the function signature, including the return type. For example:

fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
    if x.len() > y.len() {
        x
    } else {
        y
    }
}

In this function, 'a is a lifetime parameter that applies to x, y, and the return type. This means that all the references -- x, y, and the resulting string slice -- must have the same lifetime 'a. Rust will ensure that the reference returned by longest is valid for the duration of 'a, which is determined by the scope of x and y in the calling code. As a result, the lifetime of the return value is effectively the shortest of the lifetimes of x and y.

Can I avoid these problems by returned an owned string?

Yes, you can avoid dealing with lifetimes in this case by returning an owned String instead of a string slice &str. This is because an owned String is allocated on the heap and its ownership can be transferred, thus it is not tied to any particular scope or lifetime. Here's how you would modify the longest function to return a String:

fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
    if x.len() > y.len() {
        x
    } else {
        y
    }
}

In this function, 'a is a lifetime parameter that applies to x, y, and the return type. This means that all the references -- x, y, and the resulting string slice -- must have the same lifetime 'a. Rust will ensure that the reference returned by longest is valid for the duration of 'a, which is determined by the scope of x and y in the calling code. As a result, the lifetime of the return value is effectively the shortest of the lifetimes of x and y.