https://gcc.gnu.org/bugzilla/show_bug.cgi?id=61494
Richard Biener <rguenth at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Keywords| |wrong-code
Status|UNCONFIRMED |NEW
Last reconfirmed| |2014-06-13
Component|target |rtl-optimization
Ever confirmed|0 |1
--- Comment #1 from Richard Biener <rguenth at gcc dot gnu.org> ---
RTL CSE optimizes x - 0.0 to x. I thought it does nothing for FP ops ...
Confirmed.
static rtx
simplify_binary_operation_1 (enum rtx_code code, enum machine_mode mode,
rtx op0, rtx op1, rtx trueop0, rtx trueop1)
{
...
case PLUS:
/* Maybe simplify x + 0 to x. The two expressions are equivalent
when x is NaN, infinite, or finite and nonzero. They aren't
when x is -0 and the rounding mode is not towards -infinity,
since (-0) + 0 is then 0. */
if (!HONOR_SIGNED_ZEROS (mode) && trueop1 == CONST0_RTX (mode))
return op0;
...
case MINUS:
/* We can't assume x-x is 0 even with non-IEEE floating point,
but since it is zero except in very strange circumstances, we
will treat it as zero with -ffinite-math-only. */
if (rtx_equal_p (trueop0, trueop1)
&& ! side_effects_p (op0)
&& (!FLOAT_MODE_P (mode) || !HONOR_NANS (mode)))
return CONST0_RTX (mode);
/* Change subtraction from zero into negation. (0 - x) is the
same as -x when x is NaN, infinite, or finite and nonzero.
But if the mode has signed zeros, and does not round towards
-infinity, then 0 - 0 is 0, not -0. */
if (!HONOR_SIGNED_ZEROS (mode) && trueop0 == CONST0_RTX (mode))
return simplify_gen_unary (NEG, mode, op1, mode);
