rustc does not recognize a return inside a break expression for purposes of checking that every code path has the function return type

[adlerd]

I tried this code:

fn test() -> i32 {
    let loop_value = loop { break (return 5); };
}
fn main() {
    println!("{}", test());
}

I expected that this would compile. Instead rustc says (on 1.48.0 nightly 2020-09-23):

error[E0308]: mismatched types
 --> src/main.rs:1:14
  |
1 | fn test() -> i32 {
  |    ----      ^^^ expected `i32`, found `()`
  |    |
  |    implicitly returns `()` as its body has no tail or `return` expression

As a sanity check, I also tested this:

fn test() -> i32 {
    let loop_value = loop { break (return 5); };
    return 0;
}
fn main() {
    println!("{}", test());
}

This compiles and when run prints 5, indicating that that expression is indeed the place the function returns. Looking at the function it's also clearly an unconditionally reached return. Compare

fn test() -> i32 {
    let loop_value = loop { return 5; };
}
fn main() {
    println!("{}", test());
}

which also compiles and runs without error.

On the same theme, I was worried that this might erroneously typecheck:

fn test() -> i32 {
    let loop_value = loop { break (return Some(5)); };
    return 0;
}
fn main() {
    println!("{}", test());
}

but it doesn't, indicating the compiler is quite cognizant that the value returned in this position has to unify with the return type.

[varkor]

Similarly, the following doesn't work:

let loop_value = loop { break loop {} };