On Mon, Sep 25, 2023 at 3:09 PM Alexander Potapenko [off-list ref] wrote:

On Mon, Sep 25, 2023 at 2:23 PM Andy Shevchenko
[off-list ref] wrote:

quoted

On Mon, Sep 25, 2023 at 02:16:37PM +0200, Alexander Potapenko wrote:

...

quoted

quoted

+/*
+ * Test bitmap should be big enough to include the cases when start is not in
+ * the first word, and start+nbits lands in the following word.
+ */
+#define TEST_BIT_LEN (1000)

Dunno why this didn't fire previously, but CONFIG_CPU_BIG_ENDIAN=y
kernel reports mismatches here, presumably because the last quad word
ends up partially initialized.

Hmm... But if designed and used correctly it shouldn't be the issue,
and 1000, I believe, is carefully chosen to be specifically not dividable
by pow-of-2 value.

The problem manifests already right after initialization:

static void __init test_bit_len_1000(void)
{
        DECLARE_BITMAP(bitmap, TEST_BIT_LEN);
        DECLARE_BITMAP(exp_bitmap, TEST_BIT_LEN);
        memset(bitmap, 0x00, TEST_BYTE_LEN);
        memset(exp_bitmap, 0x00, TEST_BYTE_LEN);
        expect_eq_bitmap(exp_bitmap, bitmap, TEST_BIT_LEN);
}

The problem is that there's no direct analog of memset() that can be
used to initialize bitmaps on both BE and LE systems.
bitmap_zero() and bitmap_set() work by rounding up the bitmap size to
BITS_TO_LONGS(nbits), but there's no bitmap_memset() that would do the
same for an arbitrary byte pattern.
We could call memset(..., ..., BITS_TO_LONGS(TEST_BIT_LEN)), but that
would be similar to declaring a bigger bitmap and not testing the last
24 bits.

Overall, unless allocating and initializing bitmaps with size
divisible by sizeof(long), most of bitmap.c is undefined behavior, so
I don't think it makes much sense to specifically test this case here
(given that we do not extend bitmap_equal() in the patch set).

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