drm/i915: Clarify VLV PLL p1 limits

For some reason there's a sort of off by one issue with the p1 divider.
The actual p1 limits according to
VLV2_DPLL_mphy_hsdpll_frequency_table_ww6_rev1p1.xlsm is 2-3, so we should
just say that instead of saying 1-3 and avoiding the 1 via the choice of
comparison operator.

I don't know why we're using different p1 limits for intel_limits_vlv_dac
and intel_limits_vlv_hdmi, but let's preserve that for now.

Signed-off-by: Ville Syrjälä <ville.syrjala@linux.intel.com>
Reviewed-by: Mika Kuoppala <mika.kuoppala@intel.com>
Signed-off-by: Daniel Vetter <daniel.vetter@ffwll.ch>

diff --git a/drivers/gpu/drm/i915/intel_display.c b/drivers/gpu/drm/i915/intel_display.c
index ed95eb2..99c56b8 100644 (file)
--- a/drivers/gpu/drm/i915/intel_display.c
+++ b/drivers/gpu/drm/i915/intel_display.c
@@ -317,7 +317,7 @@ static const intel_limit_t intel_limits_vlv_dac = {
        .m1 = { .min = 2, .max = 3 },
        .m2 = { .min = 11, .max = 156 },
        .p = { .min = 10, .max = 30 },
-       .p1 = { .min = 1, .max = 3 },
+       .p1 = { .min = 2, .max = 3 },
        .p2 = { .dot_limit = 270000,
                .p2_slow = 2, .p2_fast = 20 },
 };
@@ -330,7 +330,7 @@ static const intel_limit_t intel_limits_vlv_hdmi = {
        .m1 = { .min = 2, .max = 3 },
        .m2 = { .min = 11, .max = 156 },
        .p = { .min = 10, .max = 30 },
-       .p1 = { .min = 2, .max = 3 },
+       .p1 = { .min = 3, .max = 3 },
        .p2 = { .dot_limit = 270000,
                .p2_slow = 2, .p2_fast = 20 },
 };
@@ -688,7 +688,7 @@ vlv_find_best_dpll(const intel_limit_t *limit, struct drm_crtc *crtc,
 
        /* based on hardware requirement, prefer smaller n to precision */
        for (clock.n = limit->n.min; clock.n <= max_n; clock.n++) {
-               for (clock.p1 = limit->p1.max; clock.p1 > limit->p1.min; clock.p1--) {
+               for (clock.p1 = limit->p1.max; clock.p1 >= limit->p1.min; clock.p1--) {
                        for (clock.p2 = limit->p2.p2_fast; clock.p2 > 0;
                             clock.p2 -= clock.p2 > 10 ? 2 : 1) {
                                clock.p = clock.p1 * clock.p2;