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> DOS > Floating point numbers |
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AsmGuru62 25 Feb 2013, 22:02
I learned here:
http://www.website.masmforum.com/tutorials/fptute/ Great article about float point processing! So, what kind of problem you're trying to solve? Most of the difficulty are conversions to/from text: TEXT -> FLOAT POINT REGISTER FLOAT POINT REGISTER -> TEXT The rest is just watching over the FPU stack so it would not overflow. Please ask your questions -- I'll try to help. |
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25 Feb 2013, 22:02 |
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A$M 26 Feb 2013, 21:05
AsmGuru62 wrote: I learned here: Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you! Thank you!!! It's too easy... A little program: Code: org 100h use16 finit fld [num3] fld [num1] fld [num2] fmul ST0, ST1 fcom ST2 fstsw ax fwait sahf jpe error jne no mov al, "Y" jmp @f no: mov al, "N" @@: mov ah, 0Eh int 10h exit: xor ax, ax int 16h ret error: mov ah, 0Eh mov al, "E" int 10h jmp exit num1 dt 1.01 num2 dt 5.5 num3 dt 5.555 It returns Y if "1.01*5.5=5.555", else returns N. |
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26 Feb 2013, 21:05 |
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AsmGuru62 26 Feb 2013, 22:24
So, it works?!
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26 Feb 2013, 22:24 |
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A$M 27 Feb 2013, 19:43
AsmGuru62 wrote: So, it works?! Works. Now, another program more complicated, but simple, which makes a circle. Please compile it and tell your opinion. I'm 13 years old and had to search for things I did not understand. That is, if there are any errors, please tell me. Anyone can copy, modify, print or do anything else. Code:
org 100h
use16
mov ax, 13h
int 10h
finit
mov ax, 0A000h
mov es, ax
next:
call calculate
mov cx, [x]
mov dx, [y]
call drawPixel
inc [degrees]
cmp [degrees], 360;ยบ
jne next
mov ax, 0
int 16h
ret
drawPixel: ; CX and DX = X and Y
add cx, 70
add dx, 70
mov ax, dx
shl dx, 6
shl ax, 8
add dx, ax
mov bx, cx
add bx, dx
mov byte[es:bx], 15
ret
calculate:
ffree st7
ffree st6
ffree st5
ffree st4
ffree st3
fild [radius]
fild [halfcircle]
fild [degrees]
fldpi
fmul st0, st1
fdiv st0, st2
fsincos
fmul st0, st4
fistp [x]
fmul st0, st3
fist [y]
ret
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Formulas ;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Degrees (x) -> Radians = x pi/180 ;;
;; Degrees (x) Radius (r) -> X location = r cos(x) ;;
;; Degrees (x) Radius (r) -> Y location = r sin(x) ;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
radius dw 25
degrees dw 0
halfcircle dw 180
x dw ?
y dw ?
EDIT: It is dotted circles if the radius is large. No big issue. The solution would be to draw lines between the points, but it is unnecessary. |
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27 Feb 2013, 19:43 |
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AsmGuru62 27 Feb 2013, 20:11
Nice code.
I see just few things: 1. The way you return to DOS is most likely incorrect. I think is the way you do it: Code: mov ax,0 int 16h ret That RET opcode may not work if you will run your program from standard DOS command line. The proper way to stop the COM program is INT 20h: Code: mov ax,0 int 16h int 20h 2. You can make it faster by only computing the pixels for 1/4 of a circle, because the other 3/4 are a symmetrically positioned pixels just with oppisite sign for X and Y. This way you need to perform only 25% of the pixels calculation. 3. You can even speed up more by not using degrees, but radians. The circle takes up 2*PI in radians, meaning that you need a 1/360th part of it to be a 1 degree. Create a variable and set it to (2*PI)/360 and use it as a step to increment your radian value going from 0 to 90, so you avoid the conversion on every step. |
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27 Feb 2013, 20:11 |
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freecrac 28 Feb 2013, 07:41
AsmGuru62 wrote: Nice code. But if we start a *.com-file under pure DOS and from the standard DOS command line, then DOS fill the PSP with an int 20h instruction and additional the stack starts with a word of 0000. So if we terminate our application only with a ret-instruction, then our programmcounter will be set to this int 20h instruction inside of the PSP. (adopted unix like ret-methode in MSDOS 2 and above?) I can not see a big differce between using only a ret-instruction for to execute the int 20h instruction inside of the PSP, or to execute an int 20h instruction that was placed as the last instruction in our codesegment. And i never read it before that there is a problem using only a ret instruction or that it is not a proper way for to stop a COM program. The only normaly conditions are that we do not have to pop the last word from the stack and we do not have to corrupt the PSP structure and additional CS normaly pointed to our codesegment. I need more information and a clarification about the difference and the statement that it is not a propper way. What does it means exactly? Dirk |
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28 Feb 2013, 07:41 |
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AsmGuru62 28 Feb 2013, 15:20
Cool! I did not know (or just forgot) that about the PSP having INT 20h opcode.
I guess both RET and INT 20h would be fine to end a COM program. |
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28 Feb 2013, 15:20 |
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baldr 28 Feb 2013, 18:41
AsmGuru62 wrote: I guess both RET and INT 20h would be fine to end a COM program. |
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28 Feb 2013, 18:41 |
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freecrac 01 Mar 2013, 08:12
Good morning and many thanks for posting an answer.
I think our last instruction in the codesegment of a COM-executacle is also highly dependent on CS, no matter if it is a ret-instruction or an int 20h-instruction, because all together have to be in one 64 KB segment with the PSP. So if we fiddiing with CS before, then we become a problem with both instructions. But if we lost the address, then we can get back the address of the currrent PSP: RBIL->inter61b.zip->INTERRUP.H Code: --------D-2162------------------------------- INT 21 - DOS 3.0+ - GET CURRENT PSP ADDRESS AH = 62h Return: BX = segment of PSP for current process Notes: this function does not use any of the DOS-internal stacks and may thus be called at any time, even during another INT 21h call the current PSP is not necessarily the caller's PSP identical to the undocumented AH=51h Dirk |
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01 Mar 2013, 08:12 |
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