flat assembler - math calculations - converting ascii in floating point

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shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 27 Jul 2013, 21:12

Maybe there are some math experts here.
I want to change a number based on 10 into floating point format based on 2.
This is for the Sinclair BASIC needed.

This is not standard format, its a bit better than single precision.

The mantissa has 32 bits and the exponent has 8 bits (with bias of 80h).
So I now can convert an integer in ascii (1234567890) into 32 bit mantissa.

If it is a floating point, it could be formatted in E-format. 12345.67890 = 1234567890E-5.

The allowed range is 10**+/-38.

What would be the best way to convert a 10-based exponent into a 2-based exponent ? If I convert it fully in binary I need 128 bit calculation but only the first 32 bits (significant bits) are really needed.

Any proposals ?

And it should run on most x86 computers produced lets say in the last 5 years.
SSE5 could handle with 128 bits as I know but this is a feature not widely available.

Thanks for any help and comments. Cool

27 Jul 2013, 21:12

tthsqe

Joined: 20 May 2009
Posts: 767

tthsqe 28 Jul 2013, 01:11

You will need to provide more details on the mantissa. Is the leading bit in the mantissa implied? Are negative numbers represented with a two's complement mantissa?

28 Jul 2013, 01:11

shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 28 Jul 2013, 11:13

Well, its quite easy.
I have no problem to convert a 10 digit integer into it.
For example:
1234567890

I go from first to last digit, multiply contents of edx with 10 and add the digit with an initial value of zero.

The exponent is 80h plus length of binary integer, so 5 is 101 which will give an exponent of 83h and after the mantissa is shifted to the left with first beginning binary one at MSB.

The integer 5 has mantissa 10100000 :00h:00h:00h=A0000000h.

As the binary integer is normalized due to shift to the left as maximum as possible, the first bit is used for positive/negative. It is reset for positive values and set for negative values.

The full floating point format is exponent + mantissa in 5 bytes:
So 5 is coded as 83 20 00 00 00 in hex bytes.
-5 would be 83 A0 00 00 00
0.2 (1/5) would be 7D 20 00 00 00 so in the other direction the exponent is subtracted, 80h -3 = 7Dh instead of 80h +3 = 83h.

There is a special coding for zero:
00 00 00 00 00 which is more a special condition than a floating point value.

It is more detailed described here (NEW ROM arithmetic):
http://www.users.waitrose.com/~thunor/mmcoyzx81/chapter17.html

This is my code for converting a positive integer in edx and write floating point via stos:

Code:

zx81_numtofp:
        or      edx,edx
        jne     @f
        xor     eax,eax
        stosb
        stosd
        ret
@@:     bsr     eax,edx
        mov     ecx,31
        sub     ecx,eax
        add     eax,81h
        stosb
        or      ecx,ecx
        je      zx81_ntf1
        shl     edx,cl
zx81_ntf1:
        and     edx,7fffffffh
        mov     eax,edx
        bswap   eax
        stosd
        ret

So the question is what is the best/fastest way to convert ascii floating point notations with big exponent base on 10 to this format based on 2 ?

Like 5.32999E+29 =5.32999*10**29 ??
As example.

28 Jul 2013, 11:13

cod3b453

Joined: 25 Aug 2004
Posts: 618

cod3b453 28 Jul 2013, 20:35

For 10 base 10 digits, ceil(log2(10^10)) is 34 so using 32bit intermediate for your conversion won't work.

Also the usual format for FP is (-1 * sign bit) * 1.fffff * base^exponent, where the ffff's are fractional bits in the mantissa. So 5 would be +1.01x2^2 or exponent=0x82 and mantissa=0x40000000. As tthsqe said, negative numbers will be a problem; also by having an explicit MSbit for the "1." you lose a bit of precision.

The conversion for 5.32999E+29 would be to take 5.32999 and convert it to float by looking at integer part - i.e. 5 = 101 = 1.01x2^2 therefore 1.3324975x2^2 (this can be done using integer shift on 64bit register).

Now take 10^29 and convert it to base 2 via log rules: log2(10^29)=29*log2(10)=96.33... and so the exponent is 98.33... . Now adjust the mantissa, to get an integer exponent, by multiplying by 2^0.33 i.e. (1.3324975*2^0.33) * 2^98.33/(2^0.33) =1.67496*2^98.

28 Jul 2013, 20:35

shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 29 Jul 2013, 17:03

cod3b453 wrote:

For 10 base 10 digits, ceil(log2(10^10)) is 34 so using 32bit intermediate for your conversion won't work.

That's why I wrote about 128 bit in the first posting.
And please look at the title of my code example:

This is my code for converting a positive integer (!) in edx and write floating point via stos

cod3b453 wrote:

Also the usual format for FP is (-1 * sign bit) * 1.fffff * base^exponent, where the ffff's are fractional bits in the mantissa. So 5 would be +1.01x2^2 or exponent=0x82 and mantissa=0x40000000. As tthsqe said, negative numbers will be a problem; also by having an explicit MSbit for the "1." you lose a bit of precision.

The difference is, that Sinclair used an exponent BIAS of 80h while standard IEEE 754 uses a BIAS of 7fh. Anyway my calculation is correct. I did not write something about standard floating point. We are talking here about a BASIC computer from the 80s.

cod3b453 wrote:

Now take 10^29 and convert it to base 2 via log rules: log2(10^29)=29*log2(10)=96.33... and so the exponent is 98.33... . Now adjust the mantissa, to get an integer exponent, by multiplying by 2^0.33 i.e. (1.3324975*2^0.33) * 2^98.33/(2^0.33) =1.67496*2^98.

Nice idea but there is no "log" in the assembler instruction set, isn't it ?
I need a program which converts a ascii floating point notation in the Sinclair floating point format (which differs from the IEEE 754 format).

29 Jul 2013, 17:03

cod3b453

Joined: 25 Aug 2004
Posts: 618

cod3b453 29 Jul 2013, 17:25

Your first post made no mention of "positive" and I read your second post as meaning you hadn't considered the negative case. Since I've never seen Sinclair floats, I had no idea that it was an actual convention.

Anyway, just because there's no log instruction doesn't mean you can't calculate it using series expansion: http://mathworld.wolfram.com/SeriesExpansion.html

EDIT: Actually log2(10) is constant so no log calculation is required, just mul.

Last edited by cod3b453 on 29 Jul 2013, 18:26; edited 1 time in total

29 Jul 2013, 17:25

tthsqe

Joined: 20 May 2009
Posts: 767

tthsqe 29 Jul 2013, 18:17

This is not complicated, with your description of the format you can do something like:

Code:

; Input:  esi address of input string
; Output  edi address to write 5 byte result

Parse:    xor  edx,edx  ; integer
          xor  ecx,ecx  ; exponent
          xor  eax,eax
          xor  ebx,ebx
.read:  lodsb
         test  al,al
           jz  Convert
          add  ecx,ebx
          cmp  al,'.'
           je  .dot
          sub  al,'0'
           js  Error
          cmp  al,'9'
           ja  Error
         imul  edx,10
          add  edx,eax
          jmp  .read
.dot:    test  ebx,ebx
          jnz  Error
          add  ebx,1
          jmp  .read

Error:   int3

Ten:     dd 10

Convert:   ; st0 = edx / 10^ecx
          sub  esp,16
          mov  [esp],edx
         fild  dword[esp]
         fld1
         fild  dword[Ten]
         test  ecx,ecx
           jz  .w3
.w1:      shr  ecx,1
          jnc  .w2
         fmul  st1,st0
.w2:     test  ecx,ecx
           jz  .w3
         fmul  st0,st0
          jmp  .w1
.w3:     fstp  st0
        fdivp  st1,st0

         fstp  tword[esp]
          mov  eax,dword[esp+4]
        movzx  ecx,word[esp+8]
          add  esp,16
          and  ecx,0x7FFFF
           jz  .zero
          add  ecx,16383+0x80+3

          btr  eax,31            ; reset for positive
        stosd
          mov  eax,ecx
        stosb
          ret


.zero:    xor  eax,eax
        stosd
        stosb
          ret

29 Jul 2013, 18:17

shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 31 Jul 2013, 13:51

Thanks for your example.
Not easy to understand for me right now as I don't used fpu instructions till now.

By the way, there should be a way to let FASM convert an ASCII floating point into single precision value ???

Quote:

Any numerical expression can also consist of single floating point value
(flat assembler does not allow any floating point operations at compilation
time) in the scientific notation, they can end with the f letter to be
recognized, otherwise they should contain at least one of the . or E
characters. So 1.0, 1E0 and 1f define the same floating point value,
while simple 1 defines an integer value.

I did not used before, maybe anyone could give an example.
Because I only need this conversion during compile time of FASM as I use a special version of FASM to generate BASIC code for Sinclair computers. There is no calculation necessary, just convert ascii float to single precision format. I think the conversion to the Sinclair format could be done easily when there is any conversion to binary float, isn't it ?

I did not use float definitions in source code till now, maybe someone could post an example. Very Happy

31 Jul 2013, 13:51

shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 31 Jul 2013, 13:58

Maybe this code example does the job:

Code:

format binary

abc     dt 1.2E37f

I have to analyze the output and compare format differences. Wink

31 Jul 2013, 13:58

shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 31 Jul 2013, 14:14

So I found the mantissa, the first 8 bytes in the picture:
So it's little endian format with 8 byte mantissa.
I just need the righter 4 bytes (marked blue) and maybe round it.

Now I have to find out about the exponent.
Cool, this build in function of FASM is exactly what I need. Cool

In my Sinclair mantissa it is coded as
10 71 DA DD which matches 90 71 DA DC before rounding and highest bit is not reset as this is done by Sinclair to mark positive values.

Description:	Mantissa
Filesize:	21.27 KB
Viewed:	19557 Time(s)

31 Jul 2013, 14:14

cod3b453

Joined: 25 Aug 2004
Posts: 618

cod3b453 31 Jul 2013, 16:20

If you only need compile time conversion then this will do it:

Code:

macro @f40 f
{
        virtual at 0

                dt f
                load q qword from 0
                load w word  from 8

        end virtual

        dd ((0x8000000000000000 or q) shr 32) ; TRUNCATION
       ;dd ((0x8000000000000000 or (q + 0x80000000)) shr 32) ; ROUNDING
        db (0x80 +((w and 0x7FFF) - 0x3FFF))
}

@f40 1.2E37f

This is taken from a generic macro I use for compile time float or fixed point conversion up to 64 bit.

31 Jul 2013, 16:20

shutdownall

Joined: 02 Apr 2010
Posts: 517
Location: Munich

shutdownall 31 Jul 2013, 19:52

Yes during compile time is sufficient. Thanks for your help. Wink

31 Jul 2013, 19:52

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