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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