bottom type as parameter or local variable, does that make sense?

[Paul Backus]

On Friday, 14 January 2022 at 22:55:37 UTC, Max Samukha wrote:
> On Friday, 14 January 2022 at 16:32:22 UTC, Paul Backus wrote:
>> That's not what the bug is here. `noreturn` is the type of 
>> expressions whose evaluation does not halt; i.e., it is a type 
>> with no values. Therefore, declaring a variable of type 
>> `noreturn` is a no-op: all it does is add a symbol to the 
>> current scope. No initialization is performed, because there 
>> is nothing to initialize.
>
> Then, should "noreturn x = noreturn.init;" fail? In my world, 
> it is equivalent to "noreturn x;". There is still an implicit 
> initialization, which should halt.

My point is, they're not equivalent.

* Execution of `noreturn x = noreturn.init;` requires evaluating 
the *expression* `noreturn.init`.
* Execution of `noreturn x;` does not require evaluating anything.

Yes, this is an exception to the general rule that `T x = T.init` 
and `T x;` are equivalent.

Why does this exception exist? Because for any type `T` other 
than `noreturn`, the expression `T.init` evaluates to a value of 
type `T`. However, the expression `noreturn.init` does not 
evaluate to a value of type `noreturn`--because, as previously 
stated, *there is no such thing as a value of type `noreturn`*.