Commit 8713d0f1 by Jeff Law Committed by Jeff Law

re PR tree-optimization/64910 (tree reassociation results in poor code)

2017-09-03  Jeff Law  <law@redhat.com>

	PR tree-optimization/64910
	* tree-ssa-reassoc.c (reassociate_bb): For bitwise binary ops,
	swap the first and last operand if the last is a constant.

	PR tree-optimization/64910
	* gcc.dg/tree-ssa/pr64910-2.c: New test.

From-SVN: r251659
parent 8d2b48ae
2017-09-04 Jeff Law <law@redhat.com>
PR tree-optimization/64910
* tree-ssa-reassoc.c (reassociate_bb): For bitwise binary ops,
swap the first and last operand if the last is a constant.
2017-09-04 Marek Polacek <polacek@redhat.com>
PR sanitizer/82072
......
2017-09-04 Jeff Law <law@redhat.com>
PR tree-optimization/64910
* gcc.dg/tree-ssa/pr64910-2.c: New test.
2017-09-04 Marek Polacek <polacek@redhat.com>
PR sanitizer/82072
......
/* { dg-do compile } */
/* { dg-options "-O2 -fdump-tree-reassoc1" } */
/* We want to make sure that we reassociate in a way that has the
constant last. With the constant last, it's more likely to result
in a bitfield test on targets with such capabilities. */
extern void boo ();
int b2b_uc (unsigned char u, unsigned char w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_us (unsigned short u, unsigned short w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_ui (unsigned int u, unsigned int w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_ul (unsigned long u, unsigned long w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_ull (unsigned long long u, unsigned long long w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_sc (signed char u, signed char w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_ss (signed short u, signed short w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_si (signed int u, signed int w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_sl (signed long u, signed long w)
{
if ((u & w) & 0x20)
boo ();
}
int b2b_sll (signed long long u, signed long long w)
{
if ((u & w) & 0x20)
boo ();
}
/* The AND of U & W should go into a temporary, when is then ANDed
with the constant.
First verify that we have the right number of ANDs between U and W. */
/* { dg-final { scan-tree-dump-times "\[uw\]_\[0-9\]+.D. \& \[uw\]_\[0-9\]+.D.;" 10 "reassoc1"} } */
/* Then verify that we have the right number of ANDS between a temporary
and the constant. */
/* { dg-final { scan-tree-dump-times "_\[0-9]+ \& 32;" 10 "reassoc1"} } */
/* Each function has one AND. It will have either a second AND or TEST. So
we can count the number of AND and TEST instructions. They must be 2X
the number of test functions in this file. */
/* { dg-final { scan-assembler-times "and|test" 20 { target { i?86-*-* x86_64-*-*} } } } */
/* Similarly on the m68k. The code for the long long tests is suboptimal,
which catch via the second pattern and xfail. */
/* { dg-final { scan-assembler-times "and|btst" 20 { target { m68k-*-* } } } } */
/* { dg-final { scan-assembler-not "or" { target { m68k-*-* } xfail { *-*-* } } } } */
......@@ -5762,6 +5762,18 @@ reassociate_bb (basic_block bb)
fprintf (dump_file,
"Width = %d was chosen for reassociation\n", width);
/* For binary bit operations, if the last operand in
OPS is a constant, move it to the front. This
helps ensure that we generate (X & Y) & C rather
than (X & C) & Y. The former will often match
a canonical bit test when we get to RTL. */
if ((rhs_code == BIT_AND_EXPR
|| rhs_code == BIT_IOR_EXPR
|| rhs_code == BIT_XOR_EXPR)
&& TREE_CODE (ops.last ()->op) == INTEGER_CST)
std::swap (*ops[0], *ops[ops_num - 1]);
if (width > 1
&& ops.length () > 3)
rewrite_expr_tree_parallel (as_a <gassign *> (stmt),
......
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