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61
Example Code / Re: Fibonacci, with bigger numbers
« Last post by debs3759 on November 27, 2023, 02:21:31 PM »
I had to google "algorism". I'm a mathematician (or was, before my memory started to fail) and had never heard the term! I thought you had mistranslated something :) Now I need to find an opportunity to play it in a Scrabble game :)
62
Example Code / Re: Fibonacci, with bigger numbers
« Last post by fredericopissarra on November 27, 2023, 12:44:45 PM »
The real problem is that fib(399) results in a 84 algarisms number:

108788617463475645289761992289049744844995705477812699099751202749393926359816304226

float has roughly 7 decimal algarisms of precision, double has 16 and long double, 19.
This means it is not possible to store an 84 algarisms (not 63!) integer in any standard floating point object (including __float128 with GCC, which has 34 algarisms of precision).

You have to use multiple precision arithmetic to do this.
63
Example Code / Re: Fibonacci, with bigger numbers
« Last post by alCoPaUL on November 26, 2023, 09:29:17 PM »
the algorithm for any math functions should preserve the integer size of the initial result and just spit out the desired result as

[2]
[4]2
[6]42
[..]...642

and do compression on the output big number, say compress([..]...642)

and you save time, extra loops and space

64
Example Code / Re: Fibonacci, with bigger numbers
« Last post by fredericopissarra on November 22, 2023, 10:01:28 AM »
Another way to find the nth number (an) in the sequence is

an = [Phi^n – (phi)^n] / Sqrt[5].
where Phi = (1 + Sqrt[5]) / 2
and phi = (1 – Sqrt[5]) / 2 (or (-1 / Phi))
The problem with these approaches is lack of "precision" for big n.
To make this work you have to deal with "multi precision arithmetic".

[]s
Fred
65
Example Code / Re: Fibonacci, with bigger numbers
« Last post by andyz74 on November 21, 2023, 10:31:03 PM »
Thank you debs, I will test this too.

But in my program, the major aspect was, how to handle the big numbers. :-)
66
Example Code / Re: Fibonacci, with bigger numbers
« Last post by debs3759 on November 21, 2023, 09:58:04 PM »
Another way to find the nth number (an) in the sequence is

an = [Phi^n – (phi)^n] / Sqrt[5].
where Phi = (1 + Sqrt[5]) / 2
and phi = (1 – Sqrt[5]) / 2 (or (-1 / Phi))
67
Example Code / Fibonacci, with bigger numbers
« Last post by andyz74 on November 21, 2023, 08:19:28 PM »
This is a mess of a code, I have to admire, but it works, as I see.
In this version, it is hardcoded for a maximum until fib(399), which has 63 numbers (or ciphers? sorry, don't know it correct in english...)
and a hardcoded maximum until 80 numbers (or ciphers? sorry, don't know it correct in english...)

This is Linux 64bit.

Code: [Select]
; fibogr.asm
;
; errechnet Fibonacci's und gibt jene aus.
; nasm -f elf64 fibogr.asm
; ld -s -o fibogr fibogr.o      "warum -s ? -s "strippt" die Debug-Infos weg. Wenn man debuggen will, lieber ohne -s"

[bits 64]

SECTION .data

_overflow_msg db '*** Series overflown.',0ah
len_overflow_msg equ $-_overflow_msg
_one db '1 '
_endl db 10
_leerz db 32
ziffer db 0

SECTION .bss
 f1array  :  resb 80
 f2array  :  resb 80
 f3array  :  resb 80



global _start ; global entry point export for ld

SECTION .text

_start:
start:


mov rax, 0
.nochmal1:
mov byte [f1array+rax],0
mov byte [f2array+rax],0
mov byte [f3array+rax],0

inc rax
cmp rax, 81
jne .nochmal1
mov byte [f1array],1
mov byte [f2array],1

mov r15, 300 ; HARDCODED : 300 fib-Zahlen machen.
; soviele fib() sollen errechnet werden...
cmp eax, 0
jz .done

dec r15
jz .done ; wenn fertig, ganz runter, Ende



call .ausgabe_f2
call .crlf

.calc_fibo:
call .kill_f3

mov r14, 0 ; r14 ist Index, wo im Array wir sind.
mov cl, 0
.alle_ziffern:
mov rax, 0
mov rbx, 0
mov r13, 0
mov al, byte [f1array+r14]
add [f3array+r14], al
mov bl, byte [f2array+r14]
add [f3array+r14], bl
mov bl, byte [f3array+r14]
jo .overflow ; Ergebnis zu groß für Datentyp?
nop
nop

inc r14
cmp r14, 80
jne .alle_ziffern

; *****  hier fertig mit rechnen *******



call .adjust_f3

call .shift_arrays

call .ausgabe_f3
call .crlf



 .done:
; call .crlf

dec r15
jnz .calc_fibo ; alle Fibos berechnet ?

mov eax, 1 ; Programmende
mov ebx, 0
int 80h

.overflow:
mov eax, 4
mov ebx, 1
mov ecx, _overflow_msg
mov edx, len_overflow_msg
int 80h

mov eax, 1
mov ebx, 1
int 80h

; *******************************************************************
; *********************** ab hier proceduren ************************
; *******************************************************************


.shift_arrays:
push rax
push rbx
push rcx
push rdx
mov rax, 0
.nochmal3:
mov ch, byte [f2array+rax]
mov byte [f1array+rax], ch

mov ch, byte [f3array+rax]
mov byte [f2array+rax], ch
inc rax
cmp rax, 80
jne .nochmal3
pop rdx
pop rcx
pop rbx
pop rax
ret


.crlf:
push rax
push rbx
push rcx
push rdx
mov eax, 4 ; schreiben
mov ebx, 1
mov ecx, _endl
mov edx, 1
int 80h
pop rdx
pop rcx
pop rbx
pop rax
ret

.leer:
push rax
push rbx
push rcx
push rdx
mov eax, 4 ; schreiben
mov ebx, 1
mov ecx, _leerz
mov edx, 1
int 80h
pop rdx
pop rcx
pop rbx
pop rax
ret


.ausgabe_f1:
push rax
push rbx
push rcx
push rdx
mov r12, 80 ; Zur Ausgabe des Arrays r12 als Index
.noch_ein_array_teil:
dec r12
mov rcx, 0
mov cl, byte [f1array+r12]
add ecx, "0"
mov byte[ziffer], cl
mov eax, 4 ; fertige "Zahl" ausgeben
mov ebx, 1
mov edx, 1 ; Länge, hier bei uns immer 1
mov ecx, ziffer
int 80h
cmp r12, 0
jne .noch_ein_array_teil
pop rdx
pop rcx
pop rbx
pop rax
ret

.ausgabe_f2:
push rax
push rbx
push rcx
push rdx
mov r12, 80 ; Zur Ausgabe des Arrays r12 als Index
.noch_ein_array_teil2:
dec r12
mov rcx, 0
mov cl, byte [f2array+r12]
add ecx, "0"
mov byte[ziffer], cl
mov eax, 4 ; fertige "Zahl" ausgeben
mov ebx, 1
mov edx, 1 ; Länge, hier bei uns immer 1
mov ecx, ziffer
int 80h
cmp r12, 0
jne .noch_ein_array_teil2
pop rdx
pop rcx
pop rbx
pop rax
ret

.ausgabe_f3:
push rax
push rbx
push rcx
push rdx
mov r13, 0  ; solange führende Nullen, später dann 1
mov r12, 80 ; Zur Ausgabe des Arrays r12 als Index
.noch_ein_array_teil3:
dec r12
mov rcx, 0
mov cl, byte [f3array+r12]
cmp r13, 0
jne .write_it ; wenn schon richtige Ziffern waren, dann hüpfen

cmp cl, 0
jne .write_it_vor ; wenn ab jetzt richtige Zahlen kommen, dann hüpfen

mov byte[ziffer], 32 ; führende Null soll Leerzeichen werden
jmp .ausgabe_leerz
.write_it_vor:
mov r13,1
.write_it:
add cl, "0"
mov byte[ziffer], cl
.ausgabe_leerz:
mov eax, 4 ; fertige "Zahl" ausgeben
mov ebx, 1
mov edx, 1 ; Länge, hier bei uns immer 1
mov ecx, ziffer
int 80h
cmp r12, 0
jne .noch_ein_array_teil3
pop rdx
pop rcx
pop rbx
pop rax
ret

.kill_f3:
push rax
mov rax, 0
.nochmal2:
mov byte [f3array+rax],0
inc rax
cmp rax, 81
jne .nochmal2
pop rax
ret


.adjust_f3:
push rax
push rbx ; Indexzähler
push rcx ; einzelner Wert
push rdx
mov rcx, 0
.nochmal4:
mov bx, 0
mov bl, byte [f3array+rcx]
cmp bl, 10 ; größer als 9  ?
jl .good
xor ax, ax
mov ax, bx
add byte [f3array+rcx+1], 1
xor rdx, rdx
mov bx, 10
div bx
; Zehner in eax, Einer in rdx
mov byte [f3array+rcx], dl
.good:
inc rcx
cmp rcx, 79
jne .nochmal4
pop rdx
pop rcx
pop rbx
pop rax
ret

68
Programming with NASM / Re: Printing of floating point values
« Last post by andyz74 on November 14, 2023, 07:15:45 PM »
Many thanks, I corrected my code now, and it does, what I want. :-)
69
Programming with NASM / Re: Printing of floating point values
« Last post by fredericopissarra on November 10, 2023, 11:38:54 AM »
Does anyone see, what is my failure, please?
Yep... Will be 0...

Let's take a look at 12.34 in bynary:
Code: [Select]
#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main( void )
{
  double d = 12.34;
  uint64_t *p = (uint64_t *)&d;

  printf( "%#" PRIx64 "\n", *p );
}
So, compiling, linking and running:
Code: [Select]
$ cc -O2 -o test test.c
$ ./test
0x4028ae147ae147ae
A floating point value is encoded as described here: https://en.wikipedia.org/wiki/Double-precision_floating-point_format
So, 100 (integer) is a double encoded as ~4.94066e-322 (almost 0). What you meant do to is:
Code: [Select]
; Input xmm0
; Ouput rax
fmult100_trunc:
  mulsd xmm0,[.m100]
  cvttsd2si rax,xmm0
  ret
.m100:
  dq  100.0         ; ".0" is necessary here.

PS: Try to avoid using "memory" variables since you have 16 integer registers and 16 xmm registers available.
PS2: This kind of code works only if the floating point value is in 'integer' range. But notice that a single precision floating point value has a range way larger than that. Take a look at this paper: https://legacy.cs.indiana.edu/~dyb/pubs/FP-Printing-PLDI96.pdf
70
Programming with NASM / Re: Printing of floating point values
« Last post by andyz74 on November 09, 2023, 04:58:44 PM »
OK, so I read a lot, and I try a lot, but unfortunately without much success.  I would be very pleased, if someone could help me further here.

I give a float variable "para_f" with the value 12.34

Then I get the integer part with

Code: [Select]
print_float:
cvttsd2si rdx, [para_f]     ;   rdx ist jetzt ganzzahlteil
ret

...and  with my own print-routine, I can print out the integer part 12.

Then I have a second part, where I try to multiply the "para_f" by 100 and afterwards print out, but I get a "0"

Code: [Select]
print_floatmul100:
movsd xmm0, [para_f] ; single-prec-float nach xmm0
mov dword [varq], 100            ; 100 nach varq und dann nach xmm1
movsd xmm1, [varq]
mulsd xmm0, xmm1   
movsd [varq], xmm0         ;
cvttsd2si rdx, [varq]
ret

I see in debugger, that xmm0 is at last point 4d2 hexadecimal, what is 1234 in decimal.  This is correct.
But then I get output zero, with my print-routine.

Does anyone see, what is my failure, please?
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