Messages in this thread |  | | From | Linus Torvalds <> | Date | Tue, 20 Mar 2018 16:08:30 -0700 | Subject | Re: detecting integer constant expressions in macros |
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On Tue, Mar 20, 2018 at 3:13 PM, Uecker, Martin <Martin.Uecker@med.uni-goettingen.de> wrote: > > here is an idea:
That's not "an idea".
That is either genius, or a seriously diseased mind.
I can't quite tell which.
> a test for integer constant expressions which returns an > integer constant expression itself which should be suitable > for passing to __builtin_choose_expr might be: > > #define ICE_P(x) (sizeof(int) == sizeof(*(1 ? ((void*)((x) * 0l)) : (int*)1)))
Ok, so I can see that (void *)((x)*0l)) turns into NULL when x is an ICE. Fine. So with a constant, we have
sizeof( 1 ? NULL : (int *) 1)
and the rule is that if one of the sides of a ternary operation with pointers is NULL, the end result is the other type (int *).
So yes, the above returns 'sizeof(int)'.
And if it is *not* an ICE that first pointer is still of type '(void *)', but it is not NULL.
And yes, the type conversion rules for a ternary op with two non-NULL pointers is different, and it now returns "void *".
So now the end result is (sizeof(*(void *)(x)), which on gcc is generally *different* from 'int'.
So I see two issues:
- "sizeof(*(void *)1)" is not necessalily well-defined. For gcc it is 1. But it could cause warnings.
- this will break the minds of everybody who ever sees that expression.
Those two issues might be fine, though.
> This also does not evaluate x itself on gcc although this is > not guaranteed by the standard. (And I haven't tried any older > gcc.)
Oh, I think it's guaranteed by the standard that 'sizeof()' doesn't evaluate the argument value, only the type.
I'm in awe of your truly marvelously disgusting hack. That is truly a work of art.
I'm sure it doesn't work or causes warnings for various reasons, but it's still a thing of beaty.
Linus
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