Accuracy falls off

[Mel Bishop]

I developed the below code to convert a number into an 8-digit
string such as would be found in a FAT. All is well until it
reaches some point. I haven't been able to find the exact point,
but when it reaches this point, it gives erronious results. For
example, 61,555,599 and 61,555,600 will return the same string,
i.e. 144 67 171 3.

According to the book, PB3.5 can digest numbers with 18-digits
of accuracy and 256 ^ 7 (the biggest I want it to hande here)
produces a 17-digit (BIG) number.

Any idea as to what's happening or what I am mis-reading?

	$lib all off
        $error all on
	color 14,1
	cls
        dim sector as shared quad

        f$ = "###,###,###,###,###,###"

	for sector = 0 to 256 ^ 4 - 1
        locate 3,10
        print;using$(f$,sector);

        y$ = quadword8$(bycopy sector)
	locate  5,10 : print;"Quad Word: ";y$

        locate  7,10
        for x = 1 to len(y$)
        print;using$("###  ",asc(mid$(y$,x,1)));
        next x
        if inkey$ = chr$(27) then exit for
        next sector

        end

function quadword8$(temp)
	tt = 256
	c8 = int(temp / (tt ^ 7)) : temp = temp - c8 * tt ^ 7
	c7 = int(temp / (tt ^ 6)) : temp = temp - c7 * tt ^ 6
	c6 = int(temp / (tt ^ 5)) : temp = temp - c6 * tt ^ 5
	c5 = int(temp / (tt ^ 4)) : temp = temp - c5 * tt ^ 4
	c4 = int(temp / (tt ^ 3)) : temp = temp - c4 * tt ^ 3
	c3 = int(temp / (tt ^ 2)) : temp = temp - c3 * tt ^ 2
	c2 = int(temp /  tt)
        c1 = temp - c2 * tt

	function = chr$(c1) + chr$(c2) + chr$(c3) + chr$(c4) + _
                   chr$(c5) + chr$(c6) + chr$(c7) + chr$(c8)
	end function

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[This message has been edited by Mel Bishop (edited January 14, 2003).]

[Paul Dixon]

Mel,
I'd guess, without trying it out, that you're using lower precision variables in the calculation of the 17 digit number.
Try DIMing all the variables in the subroutine as QUADs instead of reals in order to avoid the rounding.

Paul.


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[Bob Zale]

Mel,

Single precision floats have all the range you need, but only seven (7) significant digits. Use Double Precision for 16-digit or Extended precision for 18-digit. Also, you really should read the chapter on Floating Point in your manuals. It provides everything you need to know. Best of luck!

Bob Zale
PowerBASIC Inc.


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[Michael Mattias]

I think quadWord(temp as QUAD) might be a little easier as:

FUNCTION = MKL$(Temp \ &h0100000000) + MKL$(temp MOD &h0100000000)

(quad as string = least significant long + most significant long)

Also note you should never need floating point to work with these

And a still easier way

UNION Foo
 Q AS QUAD
 S AS STRING * 8
END UNION

DIM W AS Foo
 W.Q = temp (or other quad value)

 PRINT "string represention of temp is " & W.S

Convert back?

LSET W.S = (eight-byte string)
PRINT "Value of temp is" & STR$(W.Q)

MCM


[This message has been edited by Michael Mattias (edited January 14, 2003).]

[RM Green]

Mel,

Here's another way to do it and preserve accuracy.

$lib all off
$error all on
color 14,1
cls
dim sector as shared quad
f$ = "###,###,###,###,###,###"

for sector = 1 to 256 ^ 4 - 1
locate 3,10
print;using$(f$,sector);
c$ = ""

locate 7,10
a$ = hex$(sector)
if len(a$) < 32 then a$ = string$(32-len(a$),"0")+a$
for i% = 31 to 17 step -2
b$="&H"+mid$(a$,i%,2)
print;using$("### ",val(b$));
c$=c$+chr$(val(b$))
next i%

locate 5,10 rint "Quad String: ";c$

if inkey$ = chr$(27) then exit for

next sector

end

These are supposed to represent FAT cluster numbers? Forgive my
curiosity, but what is the purpose of the conversion and why an
8 byte string instead of 4?

Bob


[This message has been edited by RM Green (edited January 15, 2003).]

[Mel Bishop]

Originally posted by RM Green:
...Forgive my curiosity, but what is the purpose of the
conversion and why an 8 byte string instead of 4?

FAT style string. Not necessairly the FAT itself.

I am trying to read hard drive LBA sectors and, apparently, the
info packet requires an 8-byte string to read which sector. I
could be wrong but I haven't got that far to find out. I really
like experimenting, especially after I've backed up my hard
drive. (killing two birds with one stone so to speak).

I'll give your idea a try and see what happens. Thanks.


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[This message has been edited by Mel Bishop (edited January 15, 2003).]

[Mel Bishop]

$if 0
Okay, here's a complete test program I wrote to find out where
the trouble beings. Everything is fine up to it processing
16,777,216. The rounding errors occur when it tries to process
16,777,217. The number 16,777,217 falls far short of the maximum
as defined being a quad-number.

And, yes Bob, I did read the book and chapter. This program
should work.

Mr. Green, I tried your hex routine and it stopped at exactly
the same place. No luck with that one.

Anybody got any other ideas? Tnx in advance.

$endif

	$lib all off
        $error all on
	color 14,1
	cls
        dim sector as shared quad
        dim temp   as shared quad
        dim te     as shared quad
        dim back   as shared quad
        dim test1  as shared quad
        dim test2  as shared quad
        dim lp     as shared quad
        dim c1     as shared quad
        dim c2     as shared quad
        dim c3     as shared quad
        dim c4     as shared quad
        dim c5     as shared quad
        dim c6     as shared quad
        dim c7     as shared quad
        dim c8     as shared quad

        dim n1     as shared quad
        dim n2     as shared quad
        dim n3     as shared quad
        dim n4     as shared quad
        dim n5     as shared quad
        dim n6     as shared quad
        dim n7     as shared quad

        dim tt   as shared quad
        n7 = 256 ^ 7
        n6 = 256 ^ 6
        n5 = 256 ^ 5
        n4 = 256 ^ 4
        n3 = 256 ^ 3
        n2 = 256 ^ 2
        n1 = 256 ^ 1

        f$ = "###,###,###,###,###,###"		'
                                                '
	locate 2, 5 : print;"  Original:";      '
        locate 3, 5 : print;"Calculated:";      '
        lp = 16500000                           ' Preset the original number
                                                '
	do                                      '
        incr lp                                 '
        test1 = lp                              ' Substitute
        locate 2,20                             '
        print;using$(f$,test1);                 ' display original number
                                                '
        a$ = quadword8$(bycopy lp)              ' Calc the 8-bit string
                                                '
        locate 5,20                             '
        for x = 1 to len(a$)                    ' Disp 8-bit string
        print;using$("###  ",asc(mid$(a$,x,1)));'
        next x                                  '
                                                '
        test2 = back(bycopy a$)                 ' Convert string back to
        locate 3,20                             ' (?original?) number
        print;using$(f$,test2);                 '
                                                '
	if test1 <> test2 then call pause       ' Compare & pause if not a
                                                ' match.
        if inkey$ = chr$(27) then exit loop     '
	loop                                    '
        end                                     '
                                                '
sub pause                                       '
	while instat=0:wend                     '
        an$ = inkey$                            '
        if an$ = chr$(27) then end              '
        end sub                                 '
                                                '
function quadword8$(temp)                       '

	c8 = int(temp / n7) : temp = temp - c8 * n7
	c7 = int(temp / n6) : temp = temp - c7 * n6
	c6 = int(temp / n5) : temp = temp - c6 * n5
	c5 = int(temp / n4) : temp = temp - c5 * n4
	c4 = int(temp / n3) : temp = temp - c4 * n3
	c3 = int(temp / n2) : temp = temp - c3 * n2
	c2 = int(temp / n1)
        c1 = temp - c2 * 256

	function = chr$(c1) + chr$(c2) + chr$(c3) + chr$(c4) + _
                   chr$(c5) + chr$(c6) + chr$(c7) + chr$(c8)
	end function


function back(temp$)

        te = 0

	te = asc(mid$(temp$,1,1))
	te = te + asc(mid$(temp$,2,1)) * 256 ^ 1
	te = te + asc(mid$(temp$,3,1)) * 256 ^ 2
	te = te + asc(mid$(temp$,4,1)) * 256 ^ 3
	te = te + asc(mid$(temp$,5,1)) * 256 ^ 4
	te = te + asc(mid$(temp$,6,1)) * 256 ^ 5
	te = te + asc(mid$(temp$,7,1)) * 256 ^ 6
	te = te + asc(mid$(temp$,8,1)) * 256 ^ 7
	function = te

	end function

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[This message has been edited by Mel Bishop (edited January 15, 2003).]

[RM Green]

Oh, OK. You're talking about LBAs not FAT entries. I guess you
have defined a UDT for the LBA parameter buffer with the
starting LBA as a QUAD. You can just plug your LBA right in
there, no conversion is necessary.

Bob

[Bob Zale]

Originally posted by Mel Bishop:
$if 0
And, yes Bob, I did read the book and chapter. This program
should work.

No, it actually shouldn't work, and I'm sure it doesn't work. That's because the value is being rounded to a Single Precision Float as a function parameter. The chapter on Floating Point gives a pretty good description of what happens when the limits of the mantissa are exceeded. Change that Single Float param to something with a larger mantissa.

Best of luck with this!

Bob Zale
PowerBASIC Inc.


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[Michael Mattias]

Do not confuse the internal numerical accuracy of arithmetic with the string representation of operands or results.

FORMAT$, USING$ and STR$ are independent functions with their own 'accuracy' rules.

In your post, I cannot really tell if you are suffering from "bad math," "bad printed output" or both.

MCM

[Bob Zale]

Originally posted by Michael Mattias:
In your post, I cannot really tell if you are suffering from "bad math," "bad printed output" or both.

The answer is E) None of the above.

Repeat after me, Michael... {smile} Single precision floats as function parameters offer approximately 7 decimal digits of precision. That's insufficient for this application.

Regards,

Bob


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