On Wed, Nov 25, 2020 at 11:30:44AM -0800, Aditya Swarup wrote:
As I said in the other reply, sizeof does actually work here:
The question is not about sizeof() not working but rather the usage of ARRAY_SIZE()
macro in i915_drv.h with just extern declaration without size specified.
ARRAY_SIZE() is just sizeof(arr)/sizeof(*arr) with additional
shenanigans to check for misuse: when used with a pointer rather than an
array:
int b[0];
int *a = b;
or
void foo(int a[10])
In these cases ARRAY_SIZE(a) will not do what you expect and the macro
warns about it, because sizeof(a) will be sizeof(int *) instead of the
array size.
$ cat /tmp/a.c
#include <stdio.h>
#include "b.h"
int main(int argc, const char *argv[])
{
printf("%zu", sizeof(tgl_uy_revids));
return 0;
}
$ cat /tmp/b.h
#pragma once
struct i915_rev_steppings { int a; };
extern const struct i915_rev_steppings tgl_uy_revids[4];
You are specifying the size in the extern declaration which will make the ARRAY_SIZE()
macro work if used in the header else it will complain.
as it should
Lucas De Marchi
_______________________________________________
Intel-gfx mailing list
Intel-gfx@xxxxxxxxxxxxxxxxxxxxx
https://lists.freedesktop.org/mailman/listinfo/intel-gfx
