Lines Matching refs:surfaces
1953 dev_priv->surfaces[surf_index].flags);
1955 dev_priv->surfaces[surf_index].lower);
1957 dev_priv->surfaces[surf_index].upper);
1967 * For example : we allocate three contiguous surfaces ABC. If B is
1968 * freed, we suddenly need two surfaces to store A and C, which might
1990 /* make sure there is no overlap with existing surfaces */
1992 if ((dev_priv->surfaces[i].refcount != 0) &&
1993 (((new_lower >= dev_priv->surfaces[i].lower) &&
1994 (new_lower < dev_priv->surfaces[i].upper)) ||
1995 ((new_lower < dev_priv->surfaces[i].lower) &&
1996 (new_upper > dev_priv->surfaces[i].lower)))) {
2013 if ((dev_priv->surfaces[i].refcount == 1) &&
2014 (new->flags == dev_priv->surfaces[i].flags) &&
2015 (new_upper + 1 == dev_priv->surfaces[i].lower)) {
2022 dev_priv->surfaces[i].refcount++;
2023 dev_priv->surfaces[i].lower = s->lower;
2029 if ((dev_priv->surfaces[i].refcount == 1) &&
2030 (new->flags == dev_priv->surfaces[i].flags) &&
2031 (new_lower == dev_priv->surfaces[i].upper + 1)) {
2038 dev_priv->surfaces[i].refcount++;
2039 dev_priv->surfaces[i].upper = s->upper;
2047 if (dev_priv->surfaces[i].refcount == 0) {
2054 dev_priv->surfaces[i].refcount = 1;
2055 dev_priv->surfaces[i].lower = s->lower;
2056 dev_priv->surfaces[i].upper = s->upper;
2057 dev_priv->surfaces[i].flags = s->flags;
2079 if (dev_priv->surfaces[s->surface_index].
2081 dev_priv->surfaces[s->surface_index].
2084 if (dev_priv->surfaces[s->surface_index].
2086 dev_priv->surfaces[s->surface_index].
2089 dev_priv->surfaces[s->surface_index].refcount--;
2090 if (dev_priv->surfaces[s->surface_index].
2092 dev_priv->surfaces[s->surface_index].
3154 * - Free any alloced radeon surfaces.