From: sorokin@ps1.iaee.tuwien.ac.at (Zhenja)
Subject: 8-Byte integer arithmetics: SUMMARY
Organization: Inst. for General Electrics & Electronics, TU Vienna
Date: Sun, 20 Dec 1992 14:31:57 GMT

My original posting was:
>Could somebody help with code/reference to solve the following problem:
>
>I need to realize 8-Byte integer arithmetics on a 808x6.
>a) Add/sub two 8-Byte integers (signed at least);
>b) Mul/div two 4-Byte integers to give one 8-Byte.
>Actually, I have a 386 machine, so if it makes any difference I'd be happy
>with 386-specific code, provided it couldd be assembled by TASM 2.0 and/or
>understood by built-in assembler in TP 6.0 (or BC++ 2.0).
>
>I am familiar with 8086 assembler, and can (in principle) do it myself, but
>as long as I am not a professional, it will take at least 3-5 days to debug
>it properly.
>I realize that this problem is homework for the 1st year in CS, so
>references to widely known books (with answers!!) published before '92
>will do. Pieces of code would have saved a lot of time, however.
>There is big-integer arithmetic in 8x7, but the idea is to avoid
>co-processor to gain speed.
>
>Please answer to me directly. If I receive requests for summary,
>I'll prepare and publish it.
>Thanks in advance,
>Sorokin Zhenya.

First, I wish to thank everybody, who spent the time answering my silly
(as it proved to be) question. The most detailed answer (a ready-to-use
code, see below) came from Scott C. Sadow <scott%mycro.UUCP@mathcs.emory.edu>. 
My special thanks to him!

Since the solution is just a straight-forward extension of 4-byte 
arithmetics of 8086, no comments are needed.
Unfortunately, TP 6.0 and BC++ 2.0 reject any attempt to address extended
registers using "asm" directives within a program and the user have to 
assemble an object file separately. 

More interesting to me was to evaluate if there is any speed difference 
between 8-byte arithmetics in processor itself and with the use of a 
built-in 8-byte "integer" arithmetics in a math coprocessor. A cycle 
with some 600000 repetitions was used to perform one-two single operations
in both modes using TP 6.0. 
The results for 486DX-2, 50 MHz, 64K cache:
1. The time is mostly spent on pushing parameters, calling and cycling.
2. The use of math coprocessor is NOT SLOWER (and even faster in complex
   expressions, which do not require storage of intermediate results)!
   Only in operations like DIV and MOD the results of direct calculations
   are better within a factor of 1.5 but the speed difference due to different
   modes of parameter pushing is still higher.
These results may not be necessarily valid for 386/387 combination, where 
some motherboard-specific features can occur.

Merry X-mas and Happy New Year to everybody!
Frohe Weihnachten und guten Rutsch im Neuen Jahr!

The code below is a courtesy of Scott C. Sadow 
<scott%mycro.UUCP@mathcs.emory.edu>
Scott writes:

   Here is some example 80386 I put together (assembles with TASM):
---------------------------------------
	.386
	ideal
	jumps
	model	tiny
	codeseg
	ASSUME	DS:NOTHING
Start:
	;8 byte add - Num1 + Num2 = Result
	MOV	EAX,[dword Num1+0]	;handle lower 4 bytes
	ADD	EAX,[dword Num2+0]
	MOV	[dword Result+0],EAX
	MOV	EAX,[dword Num1+4]	;handle upper 4 bytes
	ADC	EAX,[dword Num2+4]
	MOV	[dword Result+4],EAX
	;8 byte subtract - Num1 - Num2 = Result
	MOV	EAX,[dword Num1+0]	;handle lower 4 bytes
	SUB	EAX,[dword Num2+0]
	MOV	[dword Result+0],EAX
	MOV	EAX,[dword Num1+4]	;handle upper 4 bytes
	SBB	EAX,[dword Num2+4]
	MOV	[dword Result+4],EAX
	;multiply - input is 2 4-byte number, output is 1 8-byte number
	MOV	EAX,[dword Num1]
	MOV	EDX,[dword Num2]
	MUL	EDX			;result is in EDX:EAX
	;divide - input is 2 4-byte numbers, output is 1 4-byte quotient and
	;						1 4-byte remainder
	MOV	EAX,[dword Num1]
	CDQ
	MOV	EBX,[dword Num2]
	DIV	EBX			;EAX = quotient, EDX = remainder
	;divide - input is 1 8-byte number and 1 4-byte number,
	;		output is 1 4-byte quotient and 1 4-byte remainder
	MOV	EAX,[dword Num1]
	MOV	EDX,[dword Num1+4]
	MOV	EBX,[dword Num2]
	DIV	EBX			;EAX = quotient, EDX = remainder
	MOV	AH,4CH
	INT	21H
Num1	db	1,2,3,4,5,6,7,8
Num2	db	1,2,3,4,5,6,7,8
Result	db	0,0,0,0,0,0,0,0
	end	Start
---------------------------------------
   Scott C. Sadow
   scott@mycro.UUCP
   ...gatech!nanovx!mycro!scott
---------------------------


Sorokin Zhenya.
Vienna, Austria.