Why does "if return" compile in Rust?

I found some strange Rust code in dtolnay’s mind-bending Rust quiz. Apparently, this is a valid Rust program (playground):

fn main() {
    if return { print!("1") } {}
}

According to the Rust docs:

The syntax of an if expression is a condition operand, followed by a consequent block, any number of else if conditions and blocks, and an optional trailing else block. The condition operands must have the boolean type.

MEDevel.com: Open-source for Healthcare and Education

Collecting and validating open-source software for healthcare, education, enterprise, development, medical imaging, medical records, and digital pathology.

Visit Medevel

To me it means that the return statement must somehow evaluate as a boolean, otherwise the code wouldn’t compile. But that explanation seems outlandish, and I suspect there must be something else going on.

So why does if return compile at all?

>Solution :

You’re looking for the never type, which is the result of the return expression.

And while it compiles, it does generate a warning:

warning: unreachable block in `if` or `while` expression
  |
2 |     if return { print!("1") } {}
  |        ---------------------- ^^ unreachable block in `if` or `while` expression
  |        |
  |        any code following this expression is unreachable