On 16/03/21 01:25PM, Marek Behún 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.be...@nic.cz> > Reviewed-by: Simon Glass <s...@chromium.org> > Reviewed-by: Heiko Schocher <h...@denx.de> > Reviewed-by: Bin Meng <bmeng...@gmail.com> > --- > drivers/core/regmap.c | 13 ++++++++++++- > 1 file changed, 12 insertions(+), 1 deletion(-) > > diff --git a/drivers/core/regmap.c b/drivers/core/regmap.c > index b51ce108c1..5d37006fff 100644 > --- a/drivers/core/regmap.c > +++ b/drivers/core/regmap.c > @@ -435,7 +435,16 @@ int regmap_raw_read(struct regmap *map, uint offset, > void *valp, size_t val_len) > > int regmap_read(struct regmap *map, uint offset, uint *valp) > { > - return regmap_raw_read(map, offset, valp, map->width); > + int res; > + > + *valp = 0; > + res = regmap_raw_read(map, offset, valp, map->width); > + if (res) > + return res; > + > + *valp = le32_to_cpu(*valp);
Looks like I'm a bit late to the party and Simon has already applied this patch. Anyway, I don't see why this is correct. regmap_raw_read() calls regmap_raw_read_range(), which calls the helpers __read_16(), __read_32() and so on. Take __read_16() for example. It takes the regmap's endianness and then based on that calls in_le16() or in_be16(). These calls translate to le16_to_cpu(__raw_readw(a)) or be16_to_cpu(__raw_readw(a)). Or the regmap is native endian in which case it is a simple readw(a). In all 3 cases the value returned is in cpu endianness. But you claim that "regmap_raw_read() and regmap_raw_write() behave as if the data given to these functions were in little-endian format". This is fine on a little endian cpu but on a big endian cpu you would reverse the byte order, no? Same for writes. > + > + return 0; > } > > static inline void __write_8(u8 *addr, const u8 *val, > @@ -546,6 +555,8 @@ int regmap_raw_write(struct regmap *map, uint offset, > const void *val, > > int regmap_write(struct regmap *map, uint offset, uint val) > { > + val = cpu_to_le32(val); > + > return regmap_raw_write(map, offset, &val, map->width); > } > > -- > 2.26.2 > -- Regards, Pratyush Yadav Texas Instruments Inc.