V30 PC6300 upgrades (NOT AGAIN!)

[Bill Mayhew]

Here are some results that I got.  I didn't feel like disassembling
my AT&T to put the V-30 back in, so I only quote the V-30 results.

Benchmarked with pi.c arctan series method (see below):

To calculate 1st 5000 digits of pi.

IBM PS/2 mod 80 SCO Xenix 2.2 cc -O:			0:15:38
IBM PS/2 mod 80 DOS 3.3 TC 1.5 large model:		0:16:02
Epson Equity III 8 MHz DOS 3.2 TC1.5 large model:	0:29:32
VAX 11/750 BSD 4.2 cc -O:				0:59:22
AT&T 6300 V-30, 8087 MSDOS 3.2 TC 1.5 large model:	1:09:02
AT&T 6300 V-30 no 8087:					same
AT&T Unix PC, 2 meg, cc -O Unix release 3.51:		1:53:18

The MSDOS results were obtained using the MKS Tooklits /bin/time
program from ksh:  /bin/time "./pi -5000 >/dev/nul"  The same form
was used for Unix/Xenix.  In Xenix, I used csh to run the command,
since ksh was not available.  Note that it appears that SCO Xenix
does not use the 80387, eventhogh one is installed.  I determined
that a 16 bit int divide requires 0.4 uS.  A double real divide
requires about 85 uS (!!).  They probably don't use the 80387 since
it maps into an interrupt that would interfere wit disk DMA if a
floating point exception occurs-- definitely not a good thing for a
multitasking environment.  IBM engineering strikes again.  I hope
to give the pi calculation a try on a AT&T Tower '386 (still a
prototype) in the next few days.

/*
** NAME
**	pi -- a program to calculate the value of pi
**
** SYNOPSIS
**        pi {-}digits {t}
**
** DESCRIPTION
**	Pi calculates the value of pi to the number of digits
**	specified. Output is to the standard output, if the
**	digits specifed is negative, output is formatted
**	for attractive display on an 80col printer.
**
**	The optional last argument "t" enables timing of the
**	computational loop. The pitime() function must be supplied
**	by the individual site implimentation.
**
**	This is an adaptation in C of the BASIC program "Apple Pi"
**	by Robert J. Bishop, published in Micro Apple (c) 1981
**	Micro Ink, Inc. That program, when run on an Apple II,
**	took 40 hours to solve pi to 1000 places. Arrays of digits
**	are used to impliment extended precision arithmatic. The
**	program solves pi using the series expansion:
**
**	      infinity                    infinity
**	      ____  16*(-1e(k+1))         ____  4*(-1e(k+1))
**	      \                           \
**	pi =   >    -------------    -     >    ------------
**	      /                           /
**	      ----  (2k-1)*5e(2k-1)       ----  (2k-1)*239e(2k-1)
**	      k = 1                       k = 1
**
** AUTHOR
**	Bryan Costales ( ...!dual!ptsfa!ski!eeg!bcx )
**	pi.c (c) 1985 Bryan Costales
**
** BUGS
**	
*/
 
 
#include <stdio.h>
 
/*
 * The maximum number of digits (plus 5 for rounding).
 */
#define MAX 100005
 
/*
 * Global variables
 */
int 	Sign, 
	Zero, 
	Size, 
 	Pass,	/* two passes */
	Exp,	/* exponent for divide */
	Divide; 
 
/*
 * These character arrays represent the digits for extended
 * precision arithmatic. (These may need to be int on some machines).
 */
#define DIGIT char
DIGIT Power[MAX], Term[MAX], Result[MAX];
 
main(argc, argv)
int  argc;
char *argv[];
{
	static int constant[3]={0,25,239};
	register int i;
	int charcount, printer = 0;
	void exit();	/* for lint */
 
	if(argc == 1)
	{
		(void)fprintf(stderr,"usage: pi {-}digits {t}\n");
		exit(1);
	}
 
	Size = atoi(argv[1]);
	if(Size < 0)
	{
		Size = -Size;
		printer++;
	}
 
	if(Size >= MAX-4 || Size <= 0)
	{
		(void)fprintf(stderr,"'digits' must be 1 thru %d.\n",MAX-5);
		exit(1);
	}
 
	if( (argc==3) && (*argv[2]=='t') )
	{
		(void)fprintf(stderr,"start computation: ");
		pitime();
	}
 
	for(Pass=1; Pass < 3; Pass++)
	{
		init();
 
		do
		{
			copy();
			Divide = Exp;
			div(Term);
			if ( Sign > 0 ) 
				add();
			if ( Sign < 0 ) 
				sub();
			Exp = Exp + 2;
			Sign *= -1;
			Divide = constant[Pass];
			div(Power);
			if (Pass == 2) 
				div(Power);
		} while( Zero != 0 );
	}
 
	if( (argc == 3) && (*argv[2] == 't') )
	{
		(void)fprintf(stderr,"end computation:   ");
		pitime();
	}
 
	if( printer )
	{
		(void)printf("\t\tThe value of pi" );
		(void)printf(" to %d decimal places\n\n\t\t",Size);
	}
 
	(void)printf("%d.",Result[0]);
	charcount = 0;
 
	for(i = 1; i <= Size; i++)
	{
		(void)printf( "%d", Result[i]);
		charcount++;
		if ( (charcount == 50) && printer )
		{
			(void)printf( "\n\t\t  " );
			charcount = 0;
		}
	}
	(void)printf( "\n" );
	return( 0 );	/* for lint */
}
 
add()
{
	register DIGIT *r, *t;
	register int sum, carry = 0;
    
	r = Result + Size;
	t = Term + Size;
 
	while( r >= Result )
	{
		sum = *r + *t + carry;
		carry = 0;
		if( sum >= 10 )
		{
			sum -= 10;
			carry++;
		}
		*r-- = sum;
		--t;
	}
}
 
sub()
{
	register DIGIT *r, *t;
	register int diff, loan = 0;		
	
	r = Result + Size;
	t = Term + Size;
 
	while( r >= Result )
	{
		diff = *r - *t - loan;
		loan =0;
		if( diff < 0 )
		{
			diff += 10;
			loan++;
		}
		*r-- = diff;
		--t;
	}
}
 
div(array)
register DIGIT *array;
{
	register DIGIT *end;
	register int quotient, residue, digit = 0;
 
	Zero = 0;
 
	for( end = array + Size; array <= end; )
	{
		digit    += *array;
		quotient =  digit / Divide;
		residue  =  digit % Divide;
 
		if((Zero !=0) || ((quotient+residue) != 0)) 
			Zero = 1;
		else 
			Zero = 0;
 
		*array++ = quotient;
		digit = 10 * residue;
	}
}
		
init()
{
	register DIGIT *p, *t, *r, *end;
 
	p = Power;
	t = Term;
	r = Result;
	end = Power+Size;
 
	while( p <= end )
	{
		*p++ = 0;
		*t++ = 0;
 
		if (Pass == 1) 
			*r++ = 0;
	}
 
	*Power = 16 / (Pass * Pass);
 
	if( Pass == 1 ) 
	{
		Divide = 5;
	}
	else
	{
		Divide = 239;
	}
 
	div(Power);
	Exp = 1;
	Sign = 3 - (2 * Pass);
}
 
copy()
{
	register DIGIT *t, *p, *end;
 
	t = Term;
	p = Power;
	end = Term + Size;
	
	while (t <= end)
	{
		*t++ = *p++;
	}
}
 
pitime()
{
	/* Print the current time \n */
 
	/* Here we just beep */
	(void)fprintf( stderr, "%c\n", 0x07 );
}