On Wed, Oct 11, 2023 at 07:14:06PM -0400, Steven Rostedt wrote:
> On Wed, 11 Oct 2023 16:33:48 -0600
> Ross Zwisler <zwisler@xxxxxxxxxx> wrote:
>
> > On Thu, Oct 05, 2023 at 09:22:33PM -0400, Steven Rostedt wrote:
> > > From: "Steven Rostedt (Google)" <rostedt@xxxxxxxxxxx>
> > >
> > > In the new addition to make sure that pointers passed to traceeval_init()
> > > and other functions that require a static array and not a dynamic one will
> > > cause the build to fail, this causes NULL pointers to fail the build too.
> > >
> > > Although keys must be filled, vals are allowed to be NULL. It was assumed
> > > that:
> > >
> > > (void *)vals == NULL ? TRACEEVAL_ARRAY_SIZE(vals))
> > >
> > > Would solve this, but it gcc was actually still giving a warning about
> > >
> > > warning: division 'sizeof (void *) / sizeof (void)' does not compute the number of array elements
> > >
> > > But now it actually fails to build with the magic check.
> > >
> > > Change TRACEEVAL_ARRAY_SIZE() to handle NULL for both keys and vals, by
> > > not only having:
> > >
> > > #define TRACEEVAL_ARRAY_SIZE(data) \
> > > ((void *)(data) == NULL ? 0 : __TRACEEVAL_ARRAY_SIZE(data))
> > >
> > > But that is not enough, as gcc still evaluates the second part, and it
> > > will fail to build. To handle this, change that to:
> > >
> > > #define __TRACEEVAL_ARRAY_SIZE(data) \
> > > ((sizeof(data) / (sizeof((data)[0])) + 0) + \
> > >
> > > The above adds " + 0" to the "sizeof((data)[0])" which quiets the warning
> > > mentioned above (the addition of zero breaks the normal pattern of finding
> > > an array size).
> > >
> > > (int)(sizeof(struct { \
> > > int:(-!!(__builtin_types_compatible_p(typeof(data), \
> > > typeof(&((data)[0]))) && \
> > > (void *)(data) != NULL)); \
> > >
> > > Added "&& (void *)(data) != NULL" that makes the above return false (zero)
> > > for a static array and NULL, which is exactly what we want.
> >
> > Don't we already know it's not NULL because of the check in
> > TRACEEVAL_ARRAY_SIZE()? Or do we really need to check for NULL in both
> > macros?
>
> Unfortunately what happens is that the compiler still checks the above. So
> if we have just:
>
> (int)(sizeof(struct { \
> int:(-!!(__builtin_types_compatible_p(typeof(data), \
> typeof(&((data)[0])))));
>
>
> Then the with NULL turns into:
>
> struct { int: -1; }
>
> and fails the compile because:
>
> __builtin_types_compatible_p(typeof(NULL), typeof(&((NULL)[0])))
>
> Returns true.
>
> So if we pair that with (void *)(data) != NULL, it will then return false
> and turns into:
>
> struct { int: 0; }
>
> Which is valid.
Sounds good. If you haven't landed this already you can add:
Reviewed-by: Ross Zwisler <zwisler@xxxxxxxxxx>
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