Re: [PATCH] regmap: move readable check before accessing regcache.

From: Mark Brown
Date: Fri Jun 18 2021 - 07:51:28 EST


On Fri, Jun 18, 2021 at 12:35:58PM +0100, Srinivas Kandagatla wrote:

> The issue that I encountered is when doing regmap_update_bits on
> a write only register. In regcache path this will not do the right
> thing as the register is not readable and driver which is using
> regmap_update_bits will never notice that it can not do a update
> bits on write only register leading to inconsistent writes and
> random hardware behavior.

Why will use of regmap_update_bits() mean that a driver will never
notice a write failure? Shouldn't remgap_update_bits() be fixed to
report any errors it isn't reporting, or the driver fixed to check
error codes? I really don't understand the issue you're trying to
report - what is "the right thing" and what makes you believe that a
driver can't do an _update_bits() on a write only but cached register?
Can you specify in concrete terms what the problem is.

> There seems to be missing checks in regcache_read() which is
> now added by moving the orignal check in _regmap_read() before
> accessing regcache.

> Cc: stable@xxxxxxxxxxxxxxx
> Fixes: 5d1729e7f02f ("regmap: Incorporate the regcache core into regmap")

Are you *sure* you've identified the actual issue here - nobody has seen
any problems with this in the past decade? Please don't just pick a
random commit for the sake of adding a Fixes tag.

> @@ -2677,6 +2677,9 @@ static int _regmap_read(struct regmap *map, unsigned int reg,
> int ret;
> void *context = _regmap_map_get_context(map);
>
> + if (!regmap_readable(map, reg))
> + return -EIO;
> +
> if (!map->cache_bypass) {
> ret = regcache_read(map, reg, val);
> if (ret == 0)
> @@ -2686,9 +2689,6 @@ static int _regmap_read(struct regmap *map, unsigned int reg,
> if (map->cache_only)
> return -EBUSY;
>
> - if (!regmap_readable(map, reg))
> - return -EIO;
> -

This puts the readability check before the cache check which will break
all drivers using the cache on write only registers.

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