Re: [Qemu-devel] [Qemu-ppc] [PATCH v2 1/4] target-ppc: Implement bcdcfsq. instruction

[David Gibson]

On Thu, Nov 24, 2016 at 02:31:21PM -0200, address@hidden wrote:
> 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

No, Richard means if you have hi_value == 1 and lo_value == 1 *after*
the divide.  This will happen if input has *decimal* value
1000000000000001.  The point is that 'i' will have the wrong value if
the low output word has any leading zeroes.

> 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~
> > 
> 

-- 
David Gibson                    | I'll have my music baroque, and my code
david AT gibson.dropbear.id.au  | minimalist, thank you.  NOT _the_ _other_
                                | _way_ _around_!
http://www.ozlabs.org/~dgibson

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