What is the use of `!` as return type in Rust? [duplicate]

Question

I have recently seen a code like this:

fn read() -> ! {
    unimplemented!()
}

fn read2() {
}

fn main() {
    read2();
    read();
}

I could not find any information about the ! as return type of fn read() anywhere so I don't have any idea what is this and what for.

The only thing I have found seems useless for me:

Using ! as a return type indicates to the Rust compiler that this function never returns

I don't understand what it does since omitting the type also says that the function returns nothing (the unit type actually).

Answer 1

Unit () is not nothing, it is a type, with one possible value also written ().

Furthermore, when a function returns unit (or "nothing" as you say), it actually returns. The Never type ! specifies that the function never returns, i.e. quits the program.

This is typically the return type of a panic macro:

let s = match i {
    1 => "one",
    2 => "two",
    _ => panic!("Error"),
}

In this example, note that ! can "take the role" of all the types. The compiler does not complain that one branch has type &str and another has type !.

For your information, here is a little history of the Never type.