Discussion Thread: DIP 1034--Add a Bottom Type (reboot)--Final Review

[Steven Schveighoffer]

Continuing discussion from feedback thread:

On 9/22/20 11:08 AM, Dennis wrote:
> On Tuesday, 22 September 2020 at 14:16:19 UTC, Steven Schveighoffer wrote:
>> ```
>> int x;
>> int bar();
>> auto foo() {
>>     cast(noreturn) (x = bar()); // same as {x = bar(); assert(0);}
>> }
>> ```
>>
>> This is odd, and I'm not understanding the reason for the cast if you 
>> aren't assigning.
> 
> This code has no reason, it's just an example of the rewrite that 
> happens with `cast(noreturn)`. It could actually be useful in generic 
> code, or when you are using a function that is noreturn but does not 
> have the correct return type.
> ```
> // someone else's code
> void panic(string msg);
> 
> // your code:
> auto f = () => cast(noreturn) panic("error");
> ```

The difference in the second example (which makes sense to me) is that 
you are returning the noreturn value.

If the first example said:

```
return cast(noreturn) (x = bar());
```

Then I understand. I thought the code was making an explicit point that 
not using the casted result still results in an assert(0).

> 
>> Also, earlier you say that you can declare a noreturn variable, but as 
>> long as you don't use it, you don't generate the assert 0. Is this 
>> example a contradiction of that?
> 
> I don't think so, the example does not declare a local variable. I could 
> rewrite it to this:
> 
> ```
> import std;
> int x;
> int bar();
> auto foo() {
>      noreturn y; // no assert here yet
>      writeln("this will get printed");
>      y = cast(noreturn) (x = bar());
>      // here it results in assert(0)
> }
> ```
> 
> The reason for this is argued by Johannes Loher here:
> https://forum.dlang.org/post/r8v8tt$14fk$1@digitalmars.com
> 
>> ```
>> /* 6 */ !(is(T == function)) || is(noreturn* : T)
>> /* 7 */ !(is(T == delegate)) || is(noreturn* : T)
>> ```
>>
>> What are these trying to say? I think these rules need some 
>> clarification or rewriting
> 
> In formal logic, "(not p) or q" is equivalent to "if p then q",

OK, I see what you are trying to do. The parentheses make it hard to see 
what it's trying to say. A comment or more specific explanation may help.

BTW, is(T == function) only works on actual function types, not function 
pointers. I think you meant function pointers.

-Steve