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

[Rasmus Villemoes]

On 02/02/2022 17.19, David Laight wrote:
> From: Kees Cook
>> Sent: 31 January 2022 20:44
>>
>> The __is_constexpr() macro is dark magic. Shed some light on it with
>> a comment to explain how and why it works.
>>
> ...
>> diff --git a/include/linux/const.h b/include/linux/const.h
>> index 435ddd72d2c4..7122d6a1f8ce 100644
>> --- a/include/linux/const.h
>> +++ b/include/linux/const.h
>> @@ -7,6 +7,30 @@
>>   * This returns a constant expression while determining if an argument is
>>   * a constant expression, most importantly without evaluating the argument.
>>   * Glory to Martin Uecker <Martin.Uecker@xxxxxxxxxxxxxxxxxxxxx>
>> + *
>> + * Details:
>> + * - sizeof() is an integer constant expression, and does not evaluate the
>> + *   value of its operand; it only examines the type of its operand.
>> + * - The results of comparing two integer constant expressions is also
>> + *   an integer constant expression.
>> + * - The use of literal "8" is to avoid warnings about unaligned pointers;
>> + *   these could otherwise just be "1"s.
>> + * - (long)(x) is used to avoid warnings about 64-bit types on 32-bit
>> + *   architectures.
>> + * - The C standard defines an "integer constant expression" as different
>> + *   from a "null pointer constant" (an integer constant 0 pointer).
>> + * - The conditional operator ("... ? ... : ...") returns the type of the
>> + *   operand that isn't a null pointer constant. This behavior is the
>> + *   central mechanism of the macro.
>> + * - If (x) is an integer constant expression, then the "* 0l" resolves it
>> + *   into a null pointer constant, which forces the conditional operator
>> + *   to return the type of the last operand: "(int *)".
>> + * - If (x) is not an integer constant expression, then the type of the
>> + *   conditional operator is from the first operand: "(void *)".
>> + * - sizeof(int) == 4 and sizeof(void) == 1.
>> + * - The ultimate comparison to "sizeof(int)" chooses between either:
>> + *     sizeof(*((int *) (8)) == sizeof(int)   (x was a constant expression)
>> + *     sizeof(*((void *)(8)) == sizeof(void)  (x was not a constant expression)
>>   */
>>  #define __is_constexpr(x) \
>>  	(sizeof(int) == sizeof(*(8 ? ((void *)((long)(x) * 0l)) : (int *)8)))
> 
> This has been making my head hurt all day.
> The above isn't really a true description - ?: doesn't work that way.
> Try the following for size:
> 
> - The conditional operator (?:) requires that both expressions have the
>   the same type (after numeric promotions).

No. Please read 6.5.15.3 for the preconditions, and 6.5.15.5 and
6.5.15.6 for the rules governing the type of the whole expression.

>   The type of the result is a compile time constant and doesn't depend on any
>   variables.

Yes, the type of any expression in C is known at compile time, and is
determined via the rules in the C standard. I wouldn't call it a
"compile time constant" though.

> - If the expressions have distinct non-NULL pointer types then they are both
>   cast to (void *) and the result has type 'void *'.

Wrong.

> - A NULL pointer can be made from any integer constant expression that
>   evaluates to 0, not just a literal 0.
> - So the type of (0 ? (void *)(x) : (int *)8) is 'int *' if (x) is zero
>   (because of the NULL) and (void *) otherwise because the pointer types
>   don't match.

That's basically how this macro works, but "So" is not warranted as it
does not follow from any of the previous, wrong, statements.

> You can test this by evaluating:
> 	sizeof *(0 ? (float *)4 : (int *)4)

That's an ill-formed conditional operator, and gcc says as much even
without any -Wall in effect.

warning: pointer type mismatch in conditional expression
    8 |  return sizeof(*(0 ? (float *)4 : (int *)4));


> This is 1 because of the implicit (void *) cast.

There is no such thing.

> I'd also delete the l from the 0l - it isn't needed.
> (Or at least use L)

That's probably true, I think it's a leftover from before the explicit
(long) cast was added, which was done to ensure the expression being
cast to (void*) wasn't a 64-bit type when void* is 32 bit. The 'l' was a
simple way to widen the expression to long in the case where x has a
type narrower than void*.

Rasmus