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* [Bug target/65166] New: [SH] use div1 to do R[n] = ((R[n] << 1) | T) - R[m]
@ 2015-02-22 18:12 olegendo at gcc dot gnu.org
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From: olegendo at gcc dot gnu.org @ 2015-02-22 18:12 UTC (permalink / raw)
To: gcc-bugs
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=65166
Bug ID: 65166
Summary: [SH] use div1 to do R[n] = ((R[n] << 1) | T) - R[m]
Product: gcc
Version: 5.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: target
Assignee: unassigned at gcc dot gnu.org
Reporter: olegendo at gcc dot gnu.org
Target: sh*-*-*
The single-step division instruction 'div1 Rm,Rn' does
R[n] = ((R[n] << 1) | T) - R[m]
if M = 0 and Q = 0.
Thus it could be utilized to do that calculation.
I've added the following pattern
;; R[n] = ((R[n] << 1) | T) - R[m]
(define_insn "*div1"
[(set (match_operand:SI 0 "arith_reg_dest" "=r")
(minus:SI (mult:SI (match_operand:SI 1 "arith_reg_operand" "0")
(const_int 2))
(match_operand:SI 2 "arith_reg_operand" "r")))
(clobber (reg:SI T_REG))]
"TARGET_SH1"
"div1 %1,%0"
[(set_attr "type" "arith")])
and it seems it is hit a few times in CSiBE.
The pattern above will produce wrong code, due to lack of M = Q = T = 0 setting
before div1. To fix that a div1 has to be prefixed with a div0u, which
eliminates the benefits over a regular 2 insn add,sub sequence.
For this to make sense the div0u insn should be eliminated if possible. This
is difficult to do, because div1 insn itself also does:
Q = (0x80000000 & R[n])
? (((R[n] << 1) | T) - R[m]) == 0
: (((R[n] << 1) | T) - R[m]) > R[n]
So effectively, both div0u and div1s have to be used as a pair.
If the T bit should be used as an input, the following patterns can be used
(also lacking div0u here):
;; R[n] = ((R[n] << 1) | T) - R[m]
(define_insn "div1"
[(set (match_operand:SI 0 "arith_reg_dest" "=r")
(minus:SI (ior:SI (mult:SI (match_operand:SI 1 "arith_reg_operand" "0")
(const_int 2))
(reg:SI T_REG))
(match_operand:SI 2 "arith_reg_operand" "r")))
(clobber (reg:SI T_REG))]
"TARGET_SH1"
"div1 %1,%0"
[(set_attr "type" "arith")])
(define_insn_and_split "*div1"
[(set (match_operand:SI 0 "arith_reg_dest")
(minus:SI (ior:SI (mult:SI (match_operand:SI 1 "arith_reg_operand")
(const_int 2))
(match_operand 2 "treg_set_expr"))
(match_operand:SI 3 "arith_reg_operand")))
(clobber (reg:SI T_REG))]
"TARGET_SH1 && can_create_pseudo_p ()"
"#"
"&& 1"
[(const_int 0)]
{
sh_treg_insns ti = sh_split_treg_set_expr (operands[2], curr_insn);
emit_insn (gen_div1 (operands[0], operands[1], operands[3]));
DONE;
})
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2015-02-22 18:12 [Bug target/65166] New: [SH] use div1 to do R[n] = ((R[n] << 1) | T) - R[m] olegendo at gcc dot gnu.org
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