Lines Matching refs:surfaces

2004 	    dev_priv->surfaces[surf_index].flags);
2006 dev_priv->surfaces[surf_index].lower);
2008 dev_priv->surfaces[surf_index].upper);
2019 * For example : we allocate three contigous surfaces ABC. If B is
2020 * freed, we suddenly need two surfaces to store A and C, which might
2041 /* make sure there is no overlap with existing surfaces */
2043 if ((dev_priv->surfaces[i].refcount != 0) &&
2044 (((new_lower >= dev_priv->surfaces[i].lower) &&
2045 (new_lower < dev_priv->surfaces[i].upper)) ||
2046 ((new_lower < dev_priv->surfaces[i].lower) &&
2047 (new_upper > dev_priv->surfaces[i].lower)))) {
2064 if ((dev_priv->surfaces[i].refcount == 1) &&
2065 (new->flags == dev_priv->surfaces[i].flags) &&
2066 (new_upper + 1 == dev_priv->surfaces[i].lower)) {
2073 dev_priv->surfaces[i].refcount++;
2074 dev_priv->surfaces[i].lower = s->lower;
2080 if ((dev_priv->surfaces[i].refcount == 1) &&
2081 (new->flags == dev_priv->surfaces[i].flags) &&
2082 (new_lower == dev_priv->surfaces[i].upper + 1)) {
2089 dev_priv->surfaces[i].refcount++;
2090 dev_priv->surfaces[i].upper = s->upper;
2098 if (dev_priv->surfaces[i].refcount == 0) {
2105 dev_priv->surfaces[i].refcount = 1;
2106 dev_priv->surfaces[i].lower = s->lower;
2107 dev_priv->surfaces[i].upper = s->upper;
2108 dev_priv->surfaces[i].flags = s->flags;
2129 if (dev_priv->surfaces[s->surface_index].
2131 dev_priv->surfaces[s->surface_index].
2134 if (dev_priv->surfaces[s->surface_index].
2136 dev_priv->surfaces[s->surface_index].
2139 dev_priv->surfaces[s->surface_index].refcount--;
2140 if (dev_priv->surfaces[s->surface_index].
2142 dev_priv->surfaces[s->surface_index].
3417 * - Free any alloced radeon surfaces.