---------------------------------------------------------------- * MultiExecBitmapAnd * * BitmapAnd node gets the bitmaps generated from BitmapIndexScan * nodes and outputs a bitmap that ANDs all input bitmaps. * * The first input bitmap is utilized to store the result of the * AND and returned to the caller. In addition, the output points * to a newly created OpStream no
| 118 | * ---------------------------------------------------------------- |
| 119 | */ |
| 120 | Node * |
| 121 | MultiExecBitmapAnd(BitmapAndState *node) |
| 122 | { |
| 123 | PlanState **bitmapplans; |
| 124 | int nplans; |
| 125 | int i; |
| 126 | bool empty = false; |
| 127 | TIDBitmap *hbm = NULL; |
| 128 | |
| 129 | /* must provide our own instrumentation support */ |
| 130 | if (node->ps.instrument) |
| 131 | InstrStartNode(node->ps.instrument); |
| 132 | |
| 133 | /* |
| 134 | * get information from the node |
| 135 | */ |
| 136 | bitmapplans = node->bitmapplans; |
| 137 | nplans = node->nplans; |
| 138 | |
| 139 | /* |
| 140 | * Scan all the subplans and AND their result bitmaps |
| 141 | */ |
| 142 | for (i = 0; i < nplans; i++) |
| 143 | { |
| 144 | PlanState *subnode = bitmapplans[i]; |
| 145 | Node *subresult = NULL; |
| 146 | |
| 147 | subresult = MultiExecProcNode(subnode); |
| 148 | |
| 149 | if (!subresult || !(IsA(subresult, TIDBitmap) || IsA(subresult, StreamBitmap))) |
| 150 | elog(ERROR, "unrecognized result from subplan"); |
| 151 | |
| 152 | /* |
| 153 | * If this is a hash bitmap, intersect it now with other hash bitmaps. |
| 154 | * If we encounter some streamed bitmaps we'll add this hash bitmap |
| 155 | * as a stream to it. |
| 156 | */ |
| 157 | if (IsA(subresult, TIDBitmap)) |
| 158 | { |
| 159 | /* first subplan that generates a hash bitmap */ |
| 160 | if (hbm == NULL) |
| 161 | hbm = (TIDBitmap *) subresult; |
| 162 | else |
| 163 | { |
| 164 | tbm_intersect(hbm, (TIDBitmap *)subresult); |
| 165 | tbm_generic_free(subresult); |
| 166 | } |
| 167 | |
| 168 | /* |
| 169 | * If at any stage we have a completely empty bitmap, we can fall out |
| 170 | * without evaluating the remaining subplans, since ANDing them can no |
| 171 | * longer change the result. (Note: the fact that indxpath.c orders |
| 172 | * the subplans by selectivity should make this case more likely to |
| 173 | * occur.) |
| 174 | */ |
| 175 | if (tbm_is_empty(hbm)) |
| 176 | { |
| 177 | empty = true; |
no test coverage detected