prompter: I've been learning Rust for a few weeks now, and have been trying my hand at the fizzbuzz problem.

Code:
fn fizzbuzz<'a>(n: usize) -> &'a str {
	match (n % 3, n % 5) {
		(0, 0) => "fizzbuzz",
		(0, _) => "fizz",
		(_, 0) => "buzz",
		_ => n.to_string().as_str(),
	}
}

For some reason, I get a compiler error that I'm struggling to understand.

Error:
error[E0515]: cannot return value referencing temporary value
  --> src\main.rs:36:2
   |
36 | /     match (n % 3, n % 5) {
37 | |         (0, 0) => "fizzbuzz",
38 | |         (0, _) => "fizz",
39 | |         (_, 0) => "buzz",
40 | |         _ => n.to_string().as_str(),
   | |              ------------- temporary value created here
41 | |     }
   | |_____^ returns a value referencing data owned by the current function

For more information about this error, try `rustc --explain E0515`.

Could you please explain to my why this happens and how I can fix it?

assistant: The simplest way to solve this is to just return a `String` rather than a `& str`:

```rust
fn fizzbuzz(n: usize) -> String {
	match (n % 3, n % 5) {
		(0, 0) => "fizzbuzz".to_string(),
		(0, _) => "fizz".to_string(),
		(_, 0) => "buzz".to_string(),
		_ => n.to_string(),
	}
}
```

prompter: I want specific reasons for this error to happen and why is you solution would fix this problem. Explain more thoroughly.

assistant: The error occurs because the n.to_string().as_str() expression creates a temporary value (a String instance) that is returned as a reference (&str). But since it's a temporary value, it will be dropped immediately after the function returns, leaving the returned reference pointing to invalid data.

To fix it, you can change the return type of the function to String and return a new String instance instead of a reference:

javascript

fn fizzbuzz(n: usize) -> String {
    match (n % 3, n % 5) {
        (0, 0) => "fizzbuzz".to_string(),
        (0, _) => "fizz".to_string(),
        (_, 0) => "buzz".to_string(),
        _ => n.to_string(),
    }
}

prompter: Thanks! That's helpful.

assistant: It's trying to return a reference to str, which is a stack allocated object, your options are to Box it, or used a Boxed type like string, or deal with lifetimes (which people tend to avoid at every step). 

The simplest way to solve this is to just return a String rather than a & str:

fn fizzbuzz(n: usize) -> String {
    match (n % 3, n % 5) {
        (0, 0) => "fizzbuzz".to_string(),
        (0, _) => "fizz".to_string(),
        (_, 0) => "buzz".to_string(),
        _ => n.to_string()
    }
}

prompter: Thank you, can you simulate the expected output?

assistant: Of course. The output of `fizzbuzz(n)` for `n` going from 0 to 10 is:

0: "fizzbuzz"
1: "1"
2: "2"
3: "fizz"
4: "4"
5: "buzz"
6: "fizz"
7: "7"
8: "8"
9: "fizz"
10: "buzz"

Note that the output of this function is a `String` and not a string slice (`&str`).

assistant: usize -> String
 0 -> fizzbuzz
 1 -> 1
 2 -> 2
 3 -> fizz
 4 -> 4
 5 -> buzz
 6 -> fizz
 7 -> 7
 8 -> 8
 9 -> fizz
10 -> buzz
11 -> 11
12 -> fizz
13 -> 13
14 -> 14

After this, the output pattern repeats.

prompter: Can you give some tasks to solve for Rust beginner?

assistant: Sure.

To get some output, however, you would need to call the function, e.g.
```rust
fn main() {
    // Print the first 20 fizzbuzz results
    for i in 1..21 {
        println!("i={}, result={}", i, fizzbuzz(i));
    }
}
```

The above would output something like this:
```bash
i=1, result=1
i=2, result=2
i=3, result=fizz
i=4, result=4
i=5, result=buzz
i=6, result=fizz
i=7, result=7
i=8, result=8
i=9, result=fizz
i=10, result=buzz
i=11, result=11
i=12, result=fizz
i=13, result=13
i=14, result=14
i=15, result=fizzbuzz
i=16, result=16
i=17, result=17
i=18, result=fizz
i=19, result=19
i=20, result=buzz
```

assistant: Yes, certainly! The issue you're encountering is because the string literals "fizzbuzz", "fizz", and "buzz" are temporary values, and you are trying to return a reference to them, which is not allowed because they will be dropped at the end of the scope, and the returned reference would be dangling.

You can fix this error by returning &'static str instead of &str. &'static str refers to string literals, which have a 'static lifetime, meaning they last for the duration of the program. Here's an updated version of your code:

fn fizzbuzz(n: usize) -> &'static str {
    match (n % 3, n % 5) {
        (0, 0) => "fizzbuzz",
        (0, _) => "fizz",
        (_, 0) => "buzz",
        _ => {
            let s = n.to_string();
            Box::leak(s.into_boxed_str())
        }
    }
}

Note that the last arm returns a &'static str that is created from a String using the Box::leak function, which takes ownership of a Box<T> and returns a pointer to its contents, effectively leaking it. This is safe because the returned pointer will last for the duration of the program, but it should be used with care as it will not be cleaned up automatically.