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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 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 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 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:
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:
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 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 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 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:
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. 

31 Jul 2013, 13:51 

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. 

31 Jul 2013, 13:58 

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. 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.


31 Jul 2013, 14:14 

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 

31 Jul 2013, 16:20 

shutdownall 31 Jul 2013, 19:52
Yes during compile time is sufficient. Thanks for your help.


31 Jul 2013, 19:52 

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