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i386: Use a new temp slot kind for splitter to floatdi<mode>2_i387_with_xmm [PR104674]

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As mentioned in the PR, the following testcase is miscompiled for similar
reasons as the already fixed PR78791 - we use SLOT_TEMP slots in various
places during expansion and during expansion we can guarantee that the
lifetime of those temporary slot doesn't overlap.  But the following
splitter uses SLOT_TEMP too and in between expansion and split1 there is
a possibility that something extends the lifetime of SLOT_TEMP created
slots across an instruction that will be split by this splitter.

The following patch fixes it by using a new temp slot kind to make sure
it doesn't reuse a SLOT_TEMP that could be live across the instruction.

2022-02-25  Jakub Jelinek  <jakub@redhat.com>

	PR target/104674
	* config/i386/i386.h (enum ix86_stack_slot): Add SLOT_FLOATxFDI_387.
	* config/i386/i386.md (splitter to floatdi<mode>2_i387_with_xmm): Use
	SLOT_FLOATxFDI_387 rather than SLOT_TEMP.

	* gcc.target/i386/pr104674.c: New test.
This commit is contained in:
Jakub Jelinek 2022-02-25 12:06:52 +01:00
parent 873b36af99
commit eabf7bbe60
3 changed files with 34 additions and 3 deletions
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1
gcc/config/i386/i386.h
View File

@ -2414,6 +2414,7 @@ enum ix86_stack_slot
SLOT_CW_FLOOR,
SLOT_CW_CEIL,
SLOT_STV_TEMP,
SLOT_FLOATxFDI_387,
MAX_386_STACK_LOCALS
};

5
gcc/config/i386/i386.md
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@ -5412,9 +5412,8 @@
&& can_create_pseudo_p ()"
[(const_int 0)]
{
emit_insn (gen_floatdi<mode>2_i387_with_xmm
(operands[0], operands[1],
assign_386_stack_local (DImode, SLOT_TEMP)));
rtx s = assign_386_stack_local (DImode, SLOT_FLOATxFDI_387);
emit_insn (gen_floatdi<mode>2_i387_with_xmm (operands[0], operands[1], s));
DONE;
})

31
gcc/testsuite/gcc.target/i386/pr104674.c Normal file
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@ -0,0 +1,31 @@
/* PR target/104674 */
/* { dg-do run { target sse2_runtime } } */
/* { dg-options "-O2 -msse2 -mfpmath=sse" } */
__attribute__((noipa)) double
bar (double x, double y)
{
return x + y;
}
__attribute__((noipa)) double
foo (long long x)
{
long long a = x / 10000000;
int b = x % 10000000;
double s = (double) a;
double n = (double) b / 1e7;
double t = s + n;
if (t == s + 1.0)
t = bar (t, s);
return t;
}
int
main ()
{
long long n = 888888;
n = n * 10000000;
if (foo (n) != 888888.0)
__builtin_abort ();
}