Return Option inside Loop - while-loop

Question

The program aims to use a loop to check if the index of a iterator variable meets certain criteria (i.g., index == 3). If find the desired index, return Some(123), else return None.
fn main() {
fn foo() -> Option<i32> {
let mut x = 5;
let mut done = false;
while !done {
x += x - 3;
if x % 5 == 0 {
done = true;
}
for (index, value) in (5..10).enumerate() {
println!("index = {} and value = {}", index, value);
if index == 3 {
return Some(123);
}
}
return None; //capture all other other possibility. So the while loop would surely return either a Some or a None
}
}
}
The compiler gives this error:
error[E0308]: mismatched types
--> <anon>:7:9
|
7 | while !done {
| ^ expected enum `std::option::Option`, found ()
|
= note: expected type `std::option::Option<i32>`
= note: found type `()`
I think the error source might be that a while loop evaluates to a (), thus it would return a () instead of Some(123). I don't know how to return a valid Some type inside a loop.

Answer 1

The value of any while true { ... } expression is always (). So the compiler expects your foo to return an Option<i32> but finds the last value in your foo body is ().
To fix this, you can add a return None outside the original while loop. You can also use the loop construct like this:
fn main() {
// run the code
foo();
fn foo() -> Option<i32> {
let mut x = 5;
loop {
x += x - 3;
for (index, value) in (5..10).enumerate() {
println!("index = {} and value = {}", index, value);
if index == 3 {
return Some(123);
}
}
if x % 5 == 0 {
return None;
}
}
}
}
The behaviour of while true { ... } statements is maybe a bit quirky and there have been a few requests to change it.