[Bug tree-optimization/57755] New: Improve fold_binary_op_with_conditional_arg

[glisse at gcc dot gnu.org]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=57755

            Bug ID: 57755
           Summary: Improve fold_binary_op_with_conditional_arg
           Product: gcc
           Version: 4.9.0
            Status: UNCONFIRMED
          Keywords: missed-optimization
          Severity: normal
          Priority: P3
         Component: tree-optimization
          Assignee: unassigned at gcc dot gnu.org
          Reporter: glisse at gcc dot gnu.org

Hello,

fold_binary_op_with_conditional_arg performs the following:

   Transform `a + (b ? x : y)' into `b ? (a + x) : (a + y)'.
   Transform, `a + (x < y)' into `(x < y) ? (a + 1) : (a + 0)'.

It gives up in this first case (arg is 'a' above):

  if (!TREE_CONSTANT (arg)
      && (TREE_SIDE_EFFECTS (arg)
          || TREE_CODE (arg) == COND_EXPR || TREE_CODE (arg) == VEC_COND_EXPR
          || TREE_CONSTANT (true_value) || TREE_CONSTANT (false_value)))
    return NULL_TREE;

and after folding both branches:

  if (!TREE_CONSTANT (arg) && !TREE_CONSTANT (lhs) && !TREE_CONSTANT (rhs))
    return NULL_TREE;

This seems suboptimal. On the one hand, for ((a<b)?a:c)*3/2+1, it distributes
the operations to a < b ? (a * 3) / 2 + 1 : (c * 3) / 2 + 1 (we can add as many
operations with constants as we want) and this isn't completely undone later
(partially with -Os, not at all with -O3). On the other hand, for
((a<2)?-1u:0)&b, it gives up instead of producing (a<2)?b:0.

We must be careful with recursions (PR55219) and with folders performing the
reverse transformations, but I think we should be able to optimize:
(((a<2)?-1:0)&((b<1)?-1:0))!=0

(obviously, the title doesn't prevent from moving this functionality to gimple)