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Re: [Qemu-devel] [Qemu-ppc] [PATCH v2 1/4] target-ppc: Implement bcdcfsq
From: |
joserz |
Subject: |
Re: [Qemu-devel] [Qemu-ppc] [PATCH v2 1/4] target-ppc: Implement bcdcfsq. instruction |
Date: |
Thu, 24 Nov 2016 14:31:21 -0200 |
User-agent: |
Mutt/1.5.24 (2015-08-30) |
Hello Richard,
Thank you for your review, please read my answer below.
On Thu, Nov 24, 2016 at 01:43:18AM +0100, Richard Henderson wrote:
> On 11/23/2016 05:21 PM, Jose Ricardo Ziviani wrote:
> >bcdcfsq.: Decimal convert from signed quadword. It is not possible
> >to convert values less than 10^31-1 or greater than -10^31-1 to be
> >represented in packed decimal format.
> >
> >Signed-off-by: Jose Ricardo Ziviani <address@hidden>
> >---
> > target-ppc/helper.h | 1 +
> > target-ppc/int_helper.c | 45
> > +++++++++++++++++++++++++++++++++++++
> > target-ppc/translate/vmx-impl.inc.c | 7 ++++++
> > 3 files changed, 53 insertions(+)
> >
> >diff --git a/target-ppc/helper.h b/target-ppc/helper.h
> >index da00f0a..87f533c 100644
> >--- a/target-ppc/helper.h
> >+++ b/target-ppc/helper.h
> >@@ -382,6 +382,7 @@ DEF_HELPER_3(bcdcfn, i32, avr, avr, i32)
> > DEF_HELPER_3(bcdctn, i32, avr, avr, i32)
> > DEF_HELPER_3(bcdcfz, i32, avr, avr, i32)
> > DEF_HELPER_3(bcdctz, i32, avr, avr, i32)
> >+DEF_HELPER_3(bcdcfsq, i32, avr, avr, i32)
> >
> > DEF_HELPER_2(xsadddp, void, env, i32)
> > DEF_HELPER_2(xssubdp, void, env, i32)
> >diff --git a/target-ppc/int_helper.c b/target-ppc/int_helper.c
> >index 8886a72..751909c 100644
> >--- a/target-ppc/int_helper.c
> >+++ b/target-ppc/int_helper.c
> >@@ -2874,6 +2874,51 @@ uint32_t helper_bcdctz(ppc_avr_t *r, ppc_avr_t *b,
> >uint32_t ps)
> > return cr;
> > }
> >
> >+uint32_t helper_bcdcfsq(ppc_avr_t *r, ppc_avr_t *b, uint32_t ps)
> >+{
> >+ int cr;
> >+ int i;
> >+ int ox_flag = 0;
> >+ uint64_t lo_value;
> >+ uint64_t hi_value;
> >+ uint64_t max = 0x38d7ea4c68000;
>
> This is at heart a decimal number, and should be written as such.
> Also, you need ULL for a 32-bit host compile.
>
OK
> >+ if (divu128(&lo_value, &hi_value, max)) {
> >+ ox_flag = 1;
> >+ } else if (lo_value >= max && hi_value == 0) {
> >+ ox_flag = 1;
> >+ }
>
> Dispense with ox_flag and set cr = CRF_SO now.
>
OK
> >+ for (i = 1; hi_value; hi_value /= 10, i++) {
> >+ bcd_put_digit(&ret, hi_value % 10, i);
> >+ }
> >+
> >+ for (; lo_value; lo_value /= 10, i++) {
> >+ bcd_put_digit(&ret, lo_value % 10, i);
> >+ }
>
> How can this possibly work? You know there are 15 digits between high and
> low, but you continue with i++?
>
> If hi_value == 1 && lo_value == 1, this should not produce 11, but
> 10000000000000001.
>
Suppose we have hi_value = lo_value = 1
after divu128 above we will have
hi_value = 744073709551617
lo_value = 18446
then, after the first for loop:
hi_value = 0
lo_value = 18446
i = 16
ret = u128 = 0x8376822234287792511
finally, after the second loop:
hi_value = 0
lo_value = 0
i = 21
ret = u128 = 0x0000000000018446744073709551617f
Which is correct, this function converted a signed quadword to bcd correctly,
using two doubleword variables:
1 << 64 | 1 = 18446744073709551617
Am I missing anything?
Thank you!
>
> r~
>