On Tue,  4 Mar 2014 19:22:54 +0200
Imre Deak <imre.d...@intel.com> wrote:

> Whenever we request a power domain it has to guarantee that all HW
> resources are enabled that are needed to access a HW register associated
> with that power domain. In case a register is on an always-on power well
> this won't result in turning on a power well, but it may require
> enabling some other HW resource. One such resource is the HSW/BDW device
> D0 state that is required for all register accesses and thus for all
> power wells/power domains.
> 
> So far the init power domain (guaranteeing access to all HW registers)
> was part of the default i9xx always-on power well, but not the HSW/BDW
> always-on power wells. Add the domain to the latter power wells too.
> 
> Atm, all the always-on power wells have noop handlers, so this doesn't
> change the functionality.
> 
> v2:
> - clarify semantics of always-on power wells (Paulo)
> 
> Signed-off-by: Imre Deak <imre.d...@intel.com>
> ---
>  drivers/gpu/drm/i915/intel_pm.c | 3 ++-
>  1 file changed, 2 insertions(+), 1 deletion(-)
> 
> diff --git a/drivers/gpu/drm/i915/intel_pm.c b/drivers/gpu/drm/i915/intel_pm.c
> index ebbd0ed..9a608f1 100644
> --- a/drivers/gpu/drm/i915/intel_pm.c
> +++ b/drivers/gpu/drm/i915/intel_pm.c
> @@ -5388,7 +5388,8 @@ EXPORT_SYMBOL_GPL(i915_release_power_well);
>  
>  #define HSW_ALWAYS_ON_POWER_DOMAINS (                        \
>       BIT(POWER_DOMAIN_PIPE_A) |                      \
> -     BIT(POWER_DOMAIN_TRANSCODER_EDP))
> +     BIT(POWER_DOMAIN_TRANSCODER_EDP) |              \
> +     BIT(POWER_DOMAIN_INIT))
>  #define HSW_DISPLAY_POWER_DOMAINS (                          \
>       (POWER_DOMAIN_MASK & ~HSW_ALWAYS_ON_POWER_DOMAINS) |    \
>       BIT(POWER_DOMAIN_INIT))

Reviewed-by: Jesse Barnes <jbar...@virtuousgeek.org>

-- 
Jesse Barnes, Intel Open Source Technology Center
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