Re: Decimal versus binary arithmetic was Re: J4 -

>>> On 4/1/2008 at 12:43 PM, in message
<if05v3503beg04nonbeci3n8eqo2vill79@4ax.com>, Clark F
Morris<cfmpublic@ns.sympatico.ca> wrote:
> On Mon, 31 Mar 2008 22:37:12 -0600, Robert <no@e.mail> wrote:
> 
><cfmpublic@ns.sympatico.ca> wrote: 

> about 'binary 
> point. Here's an 
> You may have to do this in Assembler to get the results that I am
> talking about.  Take a look at the generated instructions.  If I had
> access to an IBM mainframe I would compile a small test program to see
> what the underlying code is.

5820 9510               L    2,1296(0,9)             NUMERATOR

4E20 D1C4               CVD  2,452(0,13)             TS2=4

D203 D1C0 C02A          MVC  448(4,13),42(12)        TS2=0
SYSLIT AT +42
F060 D1C5 0004          SRP  453(7,13),4(0),0        TS2=5

5820 9518               L    2,1304(0,9)             DENOMINATOR

4E20 D1D0               CVD  2,464(0,13)             TS2=16

FDB4 D1C0 D1D3          DP   448(12,13),467(5,13)    TS2=0
TS2=19
F896 D1E0 D1C0          ZAP  480(10,13),448(7,13)    TS2=32
TS2=0
960F D1E9               OI   489(13),X'0F'           TS2=41

D202 D1D8 C02A          MVC  472(3,13),42(12)        TS2=24
SYSLIT AT +42
D204 D1DB D1E5          MVC  475(5,13),485(13)       TS2=27
TS2=37
4F20 D1D8               CVB  2,472(0,13)             TS2=24

F144 D1DB D1E0          MVO  475(5,13),480(5,13)     TS2=27
TS2=32
4F50 D1D8               CVB  5,472(0,13)             TS2=24

5C40 C004               M    4,4(0,12)               SYSLIT AT +4

1E52                    ALR  5,2

58B0 C034               L    11,52(0,12)             PBL=1

47C0 B698               BC   12,1688(0,11)           GN=16(000A70)

5A40 C000               A    4,0(0,12)               SYSLIT AT +0

GN=16    EQU  *

1222                    LTR  2,2
47B0 B6A2               BC   11,1698(0,11)           GN=17(000A7A)
5B40 C000               S    4,0(0,12)               SYSLIT AT +0
GN=17    EQU  *
9045 9520               STM  4,5,1312(9)             QUOTIENT

Frank