On Sun, Mar 7, 2021 at 12:26 PM Marek Behún marek.behun@nic.cz wrote:
There is a serious bug in regmap_read() and regmap_write() functions where an uint pointer is cast to (void *) which is then cast to (u8 *), (u16 *), (u32 *) or (u64 *), depending on register width of the map.
For example given a regmap with 16-bit register width the code int val = 0x12340000; regmap_read(map, 0, &val); only changes the lower 16 bits of val on little-endian machines. The upper 16 bits will remain 0x1234.
Nobody noticed this probably because this bug can be triggered with regmap_write() only on big-endian architectures (which are not used by many people anymore), and on little endian this bug has consequences only if register width is 8 or 16 bits and also the memory place to which regmap_read() should store it's result has non-zero upper bits, which it seems doesn't happen anywhere in U-Boot normally. CI managed to trigger this bug in unit test of dm_test_devm_regmap_field when compiled for sandbox_defconfig using LTO.
Fix this simply by taking into account that regmap_raw_read() and regmap_raw_write() behave as if the data given to these functions were in little-endian format, i.e. use cpu_to_le32() / le32_to_cpu(). In regmap_read() also zero out the space so that we don't get invalid result if regmap_raw_read() does not fill the whole object.
Signed-off-by: Marek Behún marek.behun@nic.cz Reviewed-by: Simon Glass sjg@chromium.org Reviewed-by: Heiko Schocher hs@denx.de
drivers/core/regmap.c | 13 ++++++++++++- 1 file changed, 12 insertions(+), 1 deletion(-)
Reviewed-by: Bin Meng bmeng.cn@gmail.com
Maybe we can create a test case on QEMU PPC to cover the big endian targets?
Regards, Bin