diff options
Diffstat (limited to 'gcc/tree-ssa-reassoc.c')
| -rw-r--r-- | gcc/tree-ssa-reassoc.c | 52 |
1 files changed, 26 insertions, 26 deletions
diff --git a/gcc/tree-ssa-reassoc.c b/gcc/tree-ssa-reassoc.c index 5136aee5d32..5fd5967121d 100644 --- a/gcc/tree-ssa-reassoc.c +++ b/gcc/tree-ssa-reassoc.c @@ -106,34 +106,34 @@ along with GCC; see the file COPYING3. If not see mergetmp2 = d + e and put mergetmp2 on the merge worklist. - + so merge worklist = {mergetmp, c, mergetmp2} - + Continue building binary ops of these operations until you have only one operation left on the worklist. - + So we have - + build binary op mergetmp3 = mergetmp + c - + worklist = {mergetmp2, mergetmp3} - + mergetmp4 = mergetmp2 + mergetmp3 - + worklist = {mergetmp4} - + because we have one operation left, we can now just set the original statement equal to the result of that operation. - + This will at least expose a + b and d + e to redundancy elimination as binary operations. - + For extra points, you can reuse the old statements to build the mergetmps, since you shouldn't run out. So why don't we do this? - + Because it's expensive, and rarely will help. Most trees we are reassociating have 3 or less ops. If they have 2 ops, they already will be written into a nice single binary op. If you have 3 ops, a @@ -142,15 +142,15 @@ along with GCC; see the file COPYING3. If not see mergetmp = op1 + op2 newstmt = mergetmp + op3 - + instead of mergetmp = op2 + op3 newstmt = mergetmp + op1 - + If all three are of the same rank, you can't expose them all in a single binary operator anyway, so the above is *still* the best you can do. - + Thus, this is what we do. When we have three ops left, we check to see what order to put them in, and call it a day. As a nod to vector sum reduction, we check if any of the ops are really a phi node that is a @@ -447,7 +447,7 @@ eliminate_duplicate_pair (enum tree_code opcode, { VEC_free (operand_entry_t, heap, *ops); *ops = NULL; - add_to_ops_vec (ops, fold_convert (TREE_TYPE (last->op), + add_to_ops_vec (ops, fold_convert (TREE_TYPE (last->op), integer_zero_node)); *all_done = true; } @@ -511,7 +511,7 @@ eliminate_plus_minus_pair (enum tree_code opcode, } VEC_ordered_remove (operand_entry_t, *ops, i); - add_to_ops_vec (ops, fold_convert(TREE_TYPE (oe->op), + add_to_ops_vec (ops, fold_convert(TREE_TYPE (oe->op), integer_zero_node)); VEC_ordered_remove (operand_entry_t, *ops, currindex); reassociate_stats.ops_eliminated ++; @@ -579,7 +579,7 @@ eliminate_not_pairs (enum tree_code opcode, oe->op = build_low_bits_mask (TREE_TYPE (oe->op), TYPE_PRECISION (TREE_TYPE (oe->op))); - reassociate_stats.ops_eliminated + reassociate_stats.ops_eliminated += VEC_length (operand_entry_t, *ops) - 1; VEC_free (operand_entry_t, heap, *ops); *ops = NULL; @@ -618,9 +618,9 @@ eliminate_using_constants (enum tree_code opcode, if (dump_file && (dump_flags & TDF_DETAILS)) fprintf (dump_file, "Found & 0, removing all other ops\n"); - reassociate_stats.ops_eliminated + reassociate_stats.ops_eliminated += VEC_length (operand_entry_t, *ops) - 1; - + VEC_free (operand_entry_t, heap, *ops); *ops = NULL; VEC_safe_push (operand_entry_t, heap, *ops, oelast); @@ -646,15 +646,15 @@ eliminate_using_constants (enum tree_code opcode, if (dump_file && (dump_flags & TDF_DETAILS)) fprintf (dump_file, "Found | -1, removing all other ops\n"); - reassociate_stats.ops_eliminated + reassociate_stats.ops_eliminated += VEC_length (operand_entry_t, *ops) - 1; - + VEC_free (operand_entry_t, heap, *ops); *ops = NULL; VEC_safe_push (operand_entry_t, heap, *ops, oelast); return; } - } + } else if (integer_zerop (oelast->op)) { if (VEC_length (operand_entry_t, *ops) != 1) @@ -677,8 +677,8 @@ eliminate_using_constants (enum tree_code opcode, { if (dump_file && (dump_flags & TDF_DETAILS)) fprintf (dump_file, "Found * 0, removing all other ops\n"); - - reassociate_stats.ops_eliminated + + reassociate_stats.ops_eliminated += VEC_length (operand_entry_t, *ops) - 1; VEC_free (operand_entry_t, heap, *ops); *ops = NULL; @@ -1740,14 +1740,14 @@ repropagate_negates (void) We do this top down because we don't know whether the subtract is part of a possible chain of reassociation except at the top. - + IE given d = f + g c = a + e b = c - d q = b - r k = t - q - + we want to break up k = t - q, but we won't until we've transformed q = b - r, which won't be broken up until we transform b = c - d. |