flat assembler
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> DOS > [tasm] Displaying floating point numbers in 16bit assembly? 
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al_Fazline 18 Jan 2023, 07:40
You need to convert binary integer to decimal by dividing it by 0Ah and putting remainders into the stack, until you get zero, then you unwind them into a variable (which you need to reserve some memory for), adding '0' (30h) every time, then you add decimal point and continue, this time you need to multiply the fractional part by 0Ah and append integer part of the result + '0' to the variable, until you got enough decimal digits. Then once you are done, append a $ and call int 21h with 9 in ah as you do here. 9th function requires $ as line terminator. Most other things use 0h to terminate strings but apparently not DOS.
Also why not to use fasm instead of tasm? It's kinda easier. But anyway, either will do. 

18 Jan 2023, 07:40 

DimonSoft 19 Jan 2023, 15:16
The question is about floatingpoint values though. But the principle is nearly the same, just with more corner cases if a universally usable implementation is the goal.
P.S. I guess, I’ll never stop being thankful to Tomasz for the clean and lean FASM syntax. (TASM/MASM’s one makes me feel bad each time I see their offset and ptr junk.) 

19 Jan 2023, 15:16 

al_Fazline 19 Jan 2023, 17:59
DimonSoft, yeah, I know. You can take the integer part first, convert it, and then go with the fractional part.
Or yeah, you can first take an integer part of decimal logarithm of the number, then put 0Ah to that power+1, and divide the number by it, so it's guaranteed you'll get something below 1, then you can multiply it by 0Ah and keep track of the decimal point position to print it when appropriate and continue. Or, you can multiply the number by 0Ah in power of desired number of decimal digits, then take integer part and convert it to string, keeping in mind that you'll also need to count the digit and put decimal point in certain position. 

19 Jan 2023, 17:59 

macomics 19 Jan 2023, 20:55
DimonSoft wrote: P.S. I guess, I’ll never stop being thankful to Tomasz for the clean and lean FASM syntax. fasm also knows about ptr and can apply it. Code: ; fasm1 ver. 1.73.30 format MZ stack 128 heap 0 segment @data string db 'Hello from fasm!$' segment @text entry @text:$ mov ax, @data mov ds, ax mov es, ax lea dx, byte ptr string mov ah, 9 int 33 mov ax, $4C00 int 33 to galikk: In addition to the methods described above, you can also take the format of a real number and translate it into a string using only integer commands. But it will be long and difficult, although it is possible. Format of a real 64bit number: Code: ; bitmap (63 <> 0) sqqqqqqqqqqqqmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm s  sign bit (1) q  characteristic (12) m  mantissa of a number (51) p = q  2048 ; The q value takes 12 bits. Thus, numbers from 0 to 4095 are placed in it. p  power (10=2) ; If you subtract 2048 from this value, then the p value will become signed and will be in the range from 2048 to 2047. sign v 0.314159e1 < power \ / mantissa of a number But the position of the point is already determined by the value of the power (p). Using the value of p and dividing the bits of m into an integer and a real part, you can start translating them into a string. How to do this was written to you in the 2nd answer. 

19 Jan 2023, 20:55 

revolution 20 Jan 2023, 03:31
In theory the tan function can return numbers in the range inf to +inf so the conversion to decimal representation needs to account for all possible values.
But in practice using qword floats I can't find a value that exceeds 10^16 Code: input: 1.5707963267948963 tan: 3530114321217157.5 input: 1.5707963267948966 tan: 1.633123935319537e+16 input: 1.5707963267948968 tan: 6218431163823738.0 Code: input: 5e324 tan: 5e324 

20 Jan 2023, 03:31 

DimonSoft 22 Jan 2023, 20:48
macomics wrote: fasm also knows about ptr and can apply it. I’m well aware of that and the MASM.INC macros and I’m extremely thankful to Tomasz for making the stupid nonsense optional in FASM. Thankful both as a programmer and as a university teacher. 

22 Jan 2023, 20:48 

ProMiNick 22 Jan 2023, 22:42
may be it will help to think
demo https://disk.yandex.ru/d/wDAXIae2DGcQSA https://t.me/ChatAssembler/177696 source https://t.me/ChatAssembler/177704 screenshot https://t.me/ChatAssembler/177695 I made it for myself as newton apple to call ideas, but they still not come (for start  on what value I should myltiply any float (calculated from float) to atleast extract first decimal digit. [post edited, dropbox link added] https://www.dropbox.com/s/7mlkb38i25yudhn/PJ3.zip?dl=0 Last edited by ProMiNick on 23 Jan 2023, 07:07; edited 1 time in total 

22 Jan 2023, 22:42 

al_Fazline 23 Jan 2023, 06:16
ProMiNick, that's a bad link, it does not work without registration in Telegram.


23 Jan 2023, 06:16 

ProMiNick 23 Jan 2023, 23:11
thoughts
Code: exp mantissa value realexp mantissa 3FEB 80000000 00000000 +0.00000095367431640625(1/1M) ?(3FEB) 80000000 00000000 3FEB 8637BD05 AF6C69B6 +0.000001 ?(3FEB) 8637BD05 AF6C69B6 3FF5 80000000 00000000 +0.0009765625(1/1K) ?(3FF5) 80000000 00000000 3FF5 83126E97 8D4FDF3B +0.001 ?(3FF5) 83126E97 8D4FDF3B 3FFA CCCCCCCC CCCCCCCD +0.05 ?(3FFA) CCCCCCCC CCCCCCCD 3FFB CCCCCCCC CCCCCCCD +0.1 ?(3FFB) CCCCCCCC CCCCCCCD 3FFC CCCCCCCC CCCCCCCD +0.2 ?(3FFC) CCCCCCCC CCCCCCCD 3FFD AAAAAAAA AAAAAAAA +0.33333333333333333333 ?(3FFD) AAAAAAAA AAAAAAAA 3FFD CCCCCCCC CCCCCCCD +0.4 ?(3FFD) CCCCCCCC CCCCCCCD 3FFE 80000000 00000000 +0.5 ?(3FFE) 80000000 00000000 3FFE 99999999 9999999A +0.6 ?(3FFE) 99999999 9999999A 3FFE A0000000 00000000 +0.625 ?(3FFE) A0000000 00000000 3FFE AAAAAAAA AAAAAAAA +0.66666666666666666666 ?(3FFE) AAAAAAAA AAAAAAAA 3FFE AAAAAAAA AAAAAAAB +0.66666666666666666667 ?(3FFE) AAAAAAAA AAAAAAAB 3FFE CCCCCCCC CCCCCCCD +0.8 0(403E) CCCCCCCC CCCCCCCD 3FFF 80000000 00000000 +1.0 0(403E) 00000000 00000001 4000 80000000 00000000 +2.0 0(403E) 00000000 00000002 4000 ADF85458 A2BB4A99 +2.718281828459045235 ?(4000) ADF85458 A2BB4A99 4000 C0000000 00000000 +3.0 0(403E) 00000000 00000003 4000 C90FDAA2 2168C233 +3.141592653589793238 ?(4000) C90FDAA2 2168C234 4001 80000000 00000000 +4.0 0(403E) 00000000 00000004 4001 A0000000 00000000 +5.0 0(403E) 00000000 00000005 4002 80000000 00000000 +8.0 0(403E) 00000000 00000008 4002 A0000000 00000000 +10.0 0(403E) 00000000 0000000A 4008 C8000000 00000000 +100.0 0(403E) 00000000 00000064 4008 FA000000 00000000 +1000.0 0(403E) 00000000 000003E8 4009 80000000 00000000 +1024.0 0(403E) 00000000 00000400 400B 9C400000 00000000 +5000.0 0(403E) 00000000 00001388 4012 F4240000 00000000 +1000000.0 0(403E) 00000000 00F42400 4012 F4240000 00000000 +1000000.0 0(403E) 00000000 00F42400 401C EE6B2800 00000000 +1000000000.0 0(403E) 00000000 3B9ACA00 4026 E8D4A510 00000000 +1000000000000.0 0(403E) 000000E8 D4A51000 4030 E35FA931 A0000000 +1000000000000000.0 0(403E) 00038D7E A4C68000 403A DE0B6B3A 76400000 +1000000000000000000.0 0(403E) 0DE0B6B3 A7640000 403C 96E51BAF 02CD85CC +2718281828459045235.0 0(403E) 25B946EB C0B36173 403C AE64B77E 88C92758 +3141592653589793238.0 0(403E) 2B992DDF A23249D6 403E 8AC72304 89E80000 +10000000000000000000.0 0(403E) 8AC72304 89E80000 403E AAAAAAAA AAAAAAAA +12297829382473034410.0 0(403E) AAAAAAAA AAAAAAAA 403E CCCCCCCC CCCCCCCD +14757395258967641293.0 0(403E) CCCCCCCC CCCCCCCD 403E FFFFFFFF FFFFFFFF +18446744073709551615.0 0(403E) FFFFFFFF FFFFFFFF Code: xor ecx, ecx jmp simplest_case simplest_case.zero_low_part: add [exponent], 32 xor eax, eax xchg eax, [hidwordmantissa] mov [lodwordmantissa], eax simplest_case: bsf ecx, [lodwordmantissa] jz .zero_low_part ; no infinite loop, completely zero mantissa must be handled previously jecxz .state_done mov eax, [hidwordmantissa] shrd [lodwordmantissa], eax, cl shr [hidwordmantissa], cl add [exponent], ecx .state_done: sub [exponent],$403E ; in IEEE zero is $3FFF, but we consider $3FFF+63 as zero exponent jnz .noluck ;here qword at [lodwordmantissa] can be transformed as usual integer qword to string .noluck ;multiply or divide mantissa on value from table for exponent ((2 shl 10) shl n ): +$10 +$20 +$40 +$80 and so on test [exponent], $1F ; is real exponent misaligned and required extramantissa? jz .no_extramantissa ;for rest  [lodwordmantissa],[hidwordmantissa],[extramantissa] hold mantissa could I found values for table on what multiply/divide? could this be solved the way I think? of course I don`t touch denormalized but they too fit 63 bits, could be multiplyed some value (maybe left couple bits for extramantissa) and again translation would be integer qword to string. Code: ; 7654321076543210 7654321076543210 ;$CCCCCCCCCCCCCCCD*10 shl 0 =$8000000000000000.2 (.2  no matter, rounded) ;  ; repeat period ; [][][][ ;$A3D70A3D70A3D70A*100 shl 1 =$7FFFFFFFFFFFFFFF.D0 (.D  rounded up to 80...) ; repeat period ; [][] ;$83126E978D4FDF3B645A1CAC083126E978D4FDF3B645A1CAC0*1000 shl 2 (shr 2  no matter) ;$83126E978D4FDF3B*1000 shl 2 =$7FFFFFFFFFFFFFFF.9E0 (.D  rounded up to 80...) ;$8637BD05AF6C69B6*1000000 shl 5 =$8000000000000000.5598 (bad .5 is rouded up) ;$8CBCCC096F5088CC*10^12 shl 11 =$8000000000000000.062408C (.0  no matter, rounded) ;$9ABE14CD44753B53*10^24 shl 23 =$8000000000000000.312861A5CA2633 (.3  no matter, rounded) ;same as upper _0.5 =$8000000000000000 _0.1 =$CCCCCCCCCCCCCCCD ;_0.1 *5 =_0.5 _0.01 =$A3D70A3D70A3D70A ;_0.01 *25 shl 1 =_0.5 _0.001 =$83126E978D4FDF3B ;_0.001 *125 shl 2 =_0.5 _0.000001 =$8637BD05AF6C69B6 ;_0.000001 *15625 shl 5 =_0.5 _0.000000000001 =$8CBCCC096F5088CC ;_0.000000000001 *15625 shl 11 =_0.5 macro mantissa [flt] { virtual dt flt load mantissa qword from $$ end virtual dq mantissa } ; we could hold full twords, but every hi word of tword will be wasting of space ;mantissa +1E1,+1E2,+1E3,+1E6,+1E12 same as dq _0.1,_0.01,_0.001,_0.000001,_0.000000000001 but we dont needed to explore bicycle and assembler calculate them for us mantissa +1E1,+1E2,+1E3,+1E6,+1E12,+1E24,+1E48,+1E96,+1E192; ... multiplier table (64bit is enought precision for single, double & double extended) mantissa +1E+1,+1E+2,+1E+3,+1E+6,+1E+12,+1E+24,+1E+48,+1E+96,+1E+192; ... multiplier table 

23 Jan 2023, 23:11 

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