+ else
+ {
+ tree phi;
+ unsigned int dest_idx = single_succ_edge (bb)->dest_idx;
+
+ /* BB dominates DEST. There may be many users of the PHI
+ nodes in BB. However, there is still a trivial case we
+ can handle. If the result of every PHI in BB is used
+ only by a PHI in DEST, then we can trivially merge the
+ PHI nodes from BB into DEST. */
+ for (phi = phi_nodes (bb); phi; phi = PHI_CHAIN (phi))
+ {
+ tree result = PHI_RESULT (phi);
+ use_operand_p imm_use;
+ tree use_stmt;
+
+ /* If the PHI's result is never used, then we can just
+ ignore it. */
+ if (has_zero_uses (result))
+ continue;
+
+ /* Get the single use of the result of this PHI node. */
+ if (!single_imm_use (result, &imm_use, &use_stmt)
+ || TREE_CODE (use_stmt) != PHI_NODE
+ || bb_for_stmt (use_stmt) != dest
+ || PHI_ARG_DEF (use_stmt, dest_idx) != result)
+ break;
+ }
+
+ /* If the loop above iterated through all the PHI nodes
+ in BB, then we can merge the PHIs from BB into DEST. */
+ if (!phi)
+ *current++ = bb;
+ }