glsl: Optimize pow(2, x) --> exp2(x).

On Haswell, POW takes 24 cycles, while EXP2 only takes 14.  Plus, using
POW requires putting 2.0 in a register, while EXP2 doesn't.

I believe that EXP2 will be faster than POW on basically all GPUs, so
it makes sense to optimize it.

Looking at the savage2 subset of shader-db:
total instructions in shared programs: 113225 -> 113179 (-0.04%)
instructions in affected programs:     2139 -> 2093 (-2.15%)
instances of 'math pow':               795 -> 749 (-6.14%)
instances of 'math exp':               389 -> 435 (11.8%)

Signed-off-by: Kenneth Graunke <kenneth@whitecape.org>
Reviewed-by: Matt Turner <mattst88@gmail.com>

diff --git a/src/glsl/opt_algebraic.cpp b/src/glsl/opt_algebraic.cpp
index 5e885f786483a28191a33f2700d9fd1ae6389a6b..332f0b77b87549a9ba858210211789bc63a5b4d0 100644 (file)
--- a/src/glsl/opt_algebraic.cpp
+++ b/src/glsl/opt_algebraic.cpp
@@ -87,6 +87,12 @@ is_vec_one(ir_constant *ir)
    return (ir == NULL) ? false : ir->is_one();
 }
 
+static inline bool
+is_vec_two(ir_constant *ir)
+{
+   return (ir == NULL) ? false : ir->is_value(2.0, 2);
+}
+
 static inline bool
 is_vec_negative_one(ir_constant *ir)
 {
@@ -420,6 +426,11 @@ ir_algebraic_visitor::handle_expression(ir_expression *ir)
       /* 1^x == 1 */
       if (is_vec_one(op_const[0]))
          return op_const[0];
+
+      /* pow(2,x) == exp2(x) */
+      if (is_vec_two(op_const[0]))
+         return expr(ir_unop_exp2, ir->operands[1]);
+
       break;
 
    case ir_unop_rcp: