Re: Assembly Language - Mathematics WITHOUT maths coprocessor

On Mon, 10 Oct 2005 02:11:43 GMT, "Richard Cooper"
<spamandviruses@xxxxxx> wrote:
>On Sun, 09 Oct 2005 15:35:04 -0400, ¬a\/b <al@xxx> wrote:
>
>> if you want understand it is in chapter 14 of book
>
>Yes, but with stuff like this:
>
>>>> 4. r x.
>
>I wouldn't know where to being to understand it. I mean, why would I "r
>x" when I could "q t" or "v m" or "p s" or maybe even "d e w g d x v?"
>
>Not that I understood any of the rest of it...
>
>> if "a" is an fnum than "a.sn" is the signed number in it
>> a.sn.sn is the unsigned number in it
>> so i define +-/* for "fixed point numbers"
>>
>> fnum a, b;
>>
>> a+b = a.sn+b.sn;
>> a-b = a.sn-b.sn;
>> a*b =(a.sn*b.sn).sn>>(32*precision)
>> a/b =(a.sn.sn<<(2*32*precision))/b.sn;

this should be a/b =(a.sn.sn<<(32*precision))/b.sn; ?

>Now you see, I have no idea what you just said there.
>
>Now since we're talking about fixed point numbers, I think that you're
>talking about how to add, subtract, multiply and divide them. E.g., if
>you have 32 bit integers, and the 24 most signifigant bits are the integer
>part, and the 8 least signifigant bits are the fractional part, then you
>do this for each operation:
>
>add: a + b
>subtract: a - b
>multiply: ( a * b + 1 << 7 ) >> 8
>divide: ( a << 8 + 1 << 7 ) / b

>> for see it in 10 base
>> 12.9/3.4 = 3.7[94]
>> 1290/34 = 37.[94]
>> 12.9*3.4 = 43.8[6]
>> 129 *34 = 4386

it is for me, for see if above my +-/* are right
....
>Again, I'm a bit confused, so I'll write out my own. We'll use four
>digits, one for each byte in the dword sized numbers:

>For 012.9 / 003.4, we've basically got the numbers represented as
>integers, 0129 and 0034. Now if you remember what you learned in 4th
>grade math, since both numbers are multiplied by 10, then you get the
>correct answer when you divide, which is going to be 4 in correctly
>rounded integer math. However, we don't want the correct answer, we want
>the answer times 10, so that we've still got one fractional digit. To do
>this, we just multiply the numerator by 10 before we do the division.
>That way the numerator is multiplied by 100, which is divided by the
>denominator multiplied by 10, and so the answer comes out multiplied by 10.

sorry for me,
i in my life have no "4th grade math" and no calculus course
only "Analisi Matematica 1-2" "Fisica 1-2" "Meccanica razionale"
"Geometria" "Fisica Matematica" the only corse compiuter related could
be "Teoria e applicazione delle macchine calcolatrici"

....
>> it should be ok but i don't know how to find the position of "."
>> in the translation base2^32 --> base10
>> and base10 --> base2^32
>> that serve for read-write the fixed point number in input output.
>> Has someone some idea on how it is possible find that position?
>
>I always convert to integer before displaying.
>
>Say we've got the '<< 8' style numbers. 123456.789 in this type of
>representation would be 0x01E240CA, which converted back to decimal is
>123456.789063...
>
>Since 8 bits gets us 256 possible fractional values, that's equivelent to
>2.408 decimal digits. So we'll want to output the number with either two
>or three decimal places. Usually you want to throw away a little bit of
>the number since it likely just contains rounding errors anyway, if we
>display 2 decimal places that's equivelent to discarding the 1.356 least
>signifigant bits. Normally you'd throw away more than that, but since
>we've only got 32 bits to begin with, it might be nice to keep them even
>if that last digit might be a little inaccurate.
>
>So now we need to convert the number to decimal. To do this, we first
>multiply it by 100 since we want two decimal places, this way the last two
>digits of our number are the fractional digits. Then we shift it 8 bits
>to the right to undo the '<< 8' stuff we've represented the number as. Of
>course, we add '1 < 7' to it first to round it.
>
>Now 100 in our number format is 25600 decimal, or 0x00006400 hex.
>
>multiply: (a * b + 1 << 7) >> 8
>
>(0x01E240CA * 0x00006400 + 0x00000080 ) >> 8 = 0xBC614EE8
>
>Of course, the whole event of taking "100 << 8" and then taking the answer
>">> 8" doesn't have a net effect, even with the "1 << 7" in there, so for
>this specific multiplication you can just do "a * b" and be done with it.
>
>0x01E240CA * 0x00000064 = 0x00000000BC614EE8
>
>(At this point we want it to be a 64 bit number, in case of overflow,
>although in this particular instance, the number still fits in 32 bits as
>long as it isn't signed.)
>
>Now we add 1 << 7 and then shift it right 8 bits. This has the effect of
>dividing it by 256, which we're doing to convert the number to a number
>with no fractional bits, which makes it easy to convert to decimal.
>
>0x00000000BC614EE8 + 1 << 7 = 0xBC614F68
>
>0x00000000BC614F68 >> 8 = 0x00BC614F
>
>(At this point, we're back into 32 bits again, but that's because we chose
>to discard 1.356 bits from our number, and so now our 32 bit number is a
>30.644 bit number. If we had chosen 3 decimal places instead, then we
>would have added 1.966 bits, and so at this point we'd have a 33.996 bit
>number, and so we'd have to continue doing the math in 64 bits.)
>
>(BTW, if you're wondering how I'm calculating how many bits our numbers
>are, all you do is take the base 2 log of the number. For example, to
>represent 2 digits of decimal accuracy, you do log_2(0.01) which is
>-6.6438, and so you need 6.6438 decimal places. For a more obvious
>example, to have you numbers accurate to 1/64, you calculate log_2(1/64)
>which is -8, and so you need 8 fractional bits.)
>
>Now, 0x00BC614F in decimal is 12345679, and since that is our number times
>100 (because of the step where we multiplied by 100) the last two decimal
>digits are after the decimal point, so the number is 123456.79, which is
>123456.789063 rounded to two decimal places.
>
>To convert from decimal to fixed point, you basically just do the
>reverse. With 123456.789 as input, you convert it to an integer,
>123456789, remembering that you multiplied by 1000 to make it an integer.
>Then you convert 123456789 to binary, which is 0x75BCD15. Then you shift
>that 8 bits to the left, to make it 0x000000075BCD1500 which is the
>multiplied integer in the fixed point representation. To convert it back
>to a fraction, you divide it by 1000 using the formula "(a << 8 + 1 << 7)
>/ 1000" which gives you 0x01E240CA, which is the most accurate
>representation of that number in the fixed point math.

i think it is more easy than what you say
for input something like this:
if Lstring is the string of numbers and "a" is the number
result
fnum a, b;
unum c;
char *p1, *p2;
b.sn=read(Lstring, p1);
if(*p1==".") ++p1;
else goto label;
c=read(p1, p2);
if(p1==p2) {label:;
a=b;
return a;
}
a=b<<(32*precision);
if(precision<c.len)
{k=c.len-precision;
c>>= 32*k;
}
a+=c;
return a;

for output, something like above, should be possible
.

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