Re: [PATCH] linux/const.h: Explain how __is_constexpr() works

[Nick Desaulniers]

On Wed, Feb 2, 2022 at 3:01 PM Miguel Ojeda
<miguel.ojeda.sandonis@xxxxxxxxx> wrote:
>
> On Wed, Feb 2, 2022 at 11:20 PM David Laight <David.Laight@xxxxxxxxxx> wrote:
> >
> > The type of the result depends on the type of the 2nd and 3rd arguments.
> > Not on the value of the first one.
>
> I am not talking about the first operand. The behavior of the
> conditional operator has a few cases. Since you mentioned promotions,
> it looked like you were thinking about what happens for the arithmetic
> types case, i.e.
>
> """If both the second and third operands have arithmetic type, the
> result type that would be determined by the usual arithmetic
> conversions, were they applied to those two operands, is the type of
> the result."""
>
> which could lead to thinking that the expressions need to have the
> same type as you mentioned, but that is not true, and the arithmetic
> types case is not used in the macro either. The cases used are the
> null pointer constant vs. pointer and the pointer to void vs. pointer
> to object type.
>
> > It has nothing to with the condition, the compiler is trying to 'sort out'
> > a suitable return type.
> >
> > I suspect the mismatched pointer types might even be a gcc extension.
>
> That is why I said it does not fit the constraints of the operator.
> The standard does not describe what happens in such a case.

Since this patch is a rephrasing of
https://stackoverflow.com/a/49481218, I think the relevant citation
from the C standard is below:

```
The key here is that the conditional operator returns a different type
depending on whether one of the operands is a null pointer constant
(6.5.15.6):

[...] if one operand is a null pointer constant, the result has the
type of the other operand; otherwise, one operand is a pointer to void
or a qualified version of void, in which case the result type is a
pointer to an appropriately qualified version of void.

So, if x was an integer constant expression, then the second operand
is a null pointer constant and therefore the type of the expression is
the type of the third operand, which is a pointer to int.
```

-- 
Thanks,
~Nick Desaulniers