divmod , division+modulo problem in rtl

divmod , division+modulo problem in rtl

[Pierre Mallard]

Hi,
I need someone's help to output a single operation for
an operation such as a (div a b)+(mod a b) in c
language only...

For this I did this in unsigned QI-char- op :
1)
(define_expand "udivmodqi4"
  [(set (match_operand:QI 0 "register_operand" "=r")
        	(udiv:QI (match_operand:HI 1
"register_operand" "r")
                	 (match_operand:QI 2
"register_operand"  "r")))
   (set (match_operand:QI 3 "register_operand"
"=r")(umod:QI (match_dup 1) (match_dup 2)))]
  ""
  "{
  	rtx res = gen_reg_rtx(HImode);
  	rtx opa = gen_rtx_SUBREG(QImode,res,0);
  	rtx opb = gen_rtx_SUBREG(QImode,res,1);

emit_insn(gen__udivmodqi4(res,operands[1],operands[2]));
  	emit_move_insn(operands[0],opa);
  	emit_move_insn(operands[3],opb);
  }")

(define_insn "_udivmodqi4"
  [(set (subreg:QI (match_operand:HI 0
"register_operand" "=r")0)
        (udiv:QI (match_operand:HI 1
"register_operand" "r")
                (match_operand:QI 2 "register_operand"
 "r")))
   (set (subreg:QI (match_dup 0) 1)
        (umod:QI (match_dup 1) (match_dup 2)))]
  ""
  "nop
  	divuu %0,%1,%2"
    [(set_attr "length" "2")
   (set_attr "cc" "none")])
and that's working fine ... 
But when I try to put it for an unsigned HI(short) op:
(define_expand "udivmodhi4"
  [(set (match_operand:HI 0 "register_operand" "=r")
        	(udiv:HI (match_operand:HI 1
"register_operand" "r")
                	 (match_operand:HI 2
"register_operand"  "r")))
   (set (match_operand:HI 3 "register_operand"
"=r")(umod:HI (match_dup 1) (match_dup 2)))]
  ""
  "{
      emit_move_insn (gen_rtx_REG (HImode, 24),
operands[1]);
      emit_move_insn (gen_rtx_REG (HImode, 26),
operands[2]);
     
emit_library_call_value(gen_rtx_SYMBOL_REF(Pmode,\"__udivmodhi4\"),
gen_rtx_REG (SImode, 24) , 0, SImode, 2 , gen_rtx_REG
(HImode, 24) , HImode , gen_rtx_REG (HImode, 26) ,
HImode);
      emit_move_insn (operands[0], gen_rtx_REG
(HImode, 24));
      emit_move_insn (operands[3], gen_rtx_REG
(HImode, 26));
      DONE;
  }")

The gcc output two calls to the library function
__divmodhi4 !!!! and to the define_expand then ..

So I'm wondering what's the point I missed in writing
this ...

2)And if u got any idea how a Signed Qi division can
be handle by a Qi signed division ... let me just show
u :
My goal is 
this in c :
signed char a = (signed char)b / (signed char)c
handle by
this in rtl :
(define_expand "divmodqi4"
  [(set (match_operand:QI 0 "register_operand" "=r")
        	(div:QI (match_operand:HI 1
"register_operand" "r")
                	 (match_operand:QI 2
"register_operand"  "r")))
   (set (match_operand:QI 3 "register_operand"
"=r")(mod:QI (match_dup 1) (match_dup 2)))]
  ""
  "{
  	rtx res = gen_reg_rtx(HImode);
  	rtx opa = gen_rtx_SUBREG(QImode,res,0);
  	rtx opb = gen_rtx_SUBREG(QImode,res,1);

emit_insn(gen__divmodqi4(res,operands[1],operands[2]));
  	emit_move_insn(operands[0],opa);
  	emit_move_insn(operands[3],opb);
  }")

Well just like HI is in fact ... ? 

Thanks for u're help
Pierre

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Re: divmod , division+modulo problem in rtl

[Momchil Velikov]

>>>>> "Pierre" == Pierre Mallard <pierremallard@yahoo.fr> writes:

Pierre> Hi,
Pierre> I need someone's help to output a single operation for
Pierre> an operation such as a (div a b)+(mod a b) in c
Pierre> language only...

See if this patch will help
http://gcc.gnu.org/ml/gcc-patches/2002-07/msg01207.html

Pierre> 2)And if u got any idea how a Signed Qi division can
Pierre> be handle by a Qi signed division ... let me just show
It can't, in C at least.

~velco

Re: divmod , division+modulo problem in rtl

[Pierre Mallard]

-gcc-3.0.4, c language-
Hi and thanks a lot.
I did not get any better things after having change
those few lines.
Anyway ok I just leave it now for catching div and mod
the same time..

But the problem I got here right now is the matching
of a divmodqi4 describe in a define_expand:

My word are define as being 16 bits long same as for
my char whereas my int are 32 bits long.

when I do 
char a,b,c;
c=a/b

I can't make gcc understand I need divmodqi4 and not
divmodhi4..

Gcc is always passing throw divmodhi4 and never in
divmodqi4...

I'm really disappointed think I may have missed
something cause lots of different proc implements
divmodqi4 especially avr one which as approximately
the same definition as me for the words , char and int
except it's twice less...

In my configuration divmodqi4 dosn't look as it can be
match once...

Pierre
--- Momchil Velikov <velco@fadata.bg> a écrit : >
>>>>> "Pierre" == Pierre Mallard
> <pierremallard@yahoo.fr> writes:
> 
> Pierre> Hi,
> Pierre> I need someone's help to output a single
> operation for
> Pierre> an operation such as a (div a b)+(mod a b)
> in c
> Pierre> language only...
> 
> See if this patch will help
>
http://gcc.gnu.org/ml/gcc-patches/2002-07/msg01207.html
> 
> Pierre> 2)And if u got any idea how a Signed Qi
> division can
> Pierre> be handle by a Qi signed division ... let me
> just show
> It can't, in C at least.
> 
> ~velco
>  

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Re: divmod , division+modulo problem in rtl

[Pierre Mallard]

-gcc-3.0.4, c language-
Hi and thanks a lot.
I did not get any better things after having change
those few lines.
Anyway ok I just leave it now for catching div and mod
the same time..

But the problem I got here right now is the matching
of a divmodqi4 describe in a define_expand:

My word are define as being 16 bits long same as for
my char whereas my int are 32 bits long.

when I do 
char a,b,c;
c=a/b

I can't make gcc understand I need divmodqi4 and not
divmodhi4..

Gcc is always passing throw divmodhi4 and never in
divmodqi4...

I'm really disappointed think I may have missed
something cause lots of different proc implements
divmodqi4 especially avr one which as approximately
the same definition as me for the words , char and int
except it's twice less...

In my configuration divmodqi4 dosn't look as it can be
match once...

Pierre
--- Momchil Velikov <velco@fadata.bg> a écrit : >
>>>>> "Pierre" == Pierre Mallard
> <pierremallard@yahoo.fr> writes:
> 
> Pierre> Hi,
> Pierre> I need someone's help to output a single
> operation for
> Pierre> an operation such as a (div a b)+(mod a b)
> in c
> Pierre> language only...
> 
> See if this patch will help
>
http://gcc.gnu.org/ml/gcc-patches/2002-07/msg01207.html
> 
> Pierre> 2)And if u got any idea how a Signed Qi
> division can
> Pierre> be handle by a Qi signed division ... let me
> just show
> It can't, in C at least.
> 
> ~velco
>  

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Char as long as one Word , Int twice a Word

[Pierre Mallard]

I've tryed to define a int as long as one word on my
proc, and then automatically char is well computed...
So if my register are 16 bits long I'm forced to
define int as 16 and not 32 as I wanted first, to
solve the problem of having my char div being computed
as an int div...

Do u know if it's possible to turn of this behaviour
in gcc (make a signed char op being cast in int op)
when int are twice a register ?

 --- Momchil Velikov <velco@fadata.bg> a écrit : >
>>>>> "Pierre" == Pierre Mallard
> <pierremallard@yahoo.fr> writes:
> 
> Pierre> Hi,
> Pierre> I need someone's help to output a single
> operation for
> Pierre> an operation such as a (div a b)+(mod a b)
> in c
> Pierre> language only...
> 
> See if this patch will help
>
http://gcc.gnu.org/ml/gcc-patches/2002-07/msg01207.html
> 
> Pierre> 2)And if u got any idea how a Signed Qi
> division can
> Pierre> be handle by a Qi signed division ... let me
> just show
> It can't, in C at least.
> 
> ~velco
>  

___________________________________________________________
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