Author Topic: Using idiv  (Read 826 times)

Offline JohnG

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Using idiv
« on: January 13, 2022, 11:26:27 PM »
Hi all,

It seems simple enough, but what am I doing wrong, just trying things at the moment with no luck.

I did read somewhere that the divisor had to be a smaller data size than the dividend ?

start:
        and     rsp,    -16  ;stack align
       
       
        mov     rax,    [a]    ; a = 100  dividend
        xor     rdx,    rdx
        mov     r8,    [sum]  ; sum = 4
        idiv    dword[r8]       
        mov     rcx,    fmt
        mov     rdx,    [r8]   ; value to print
        sub     rsp,    32      ; shadow space
        call    printf
        add     rsp,    32

I may be stomping on registers ?

John

Offline Frank Kotler

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Re: Using idiv
« Reply #1 on: January 13, 2022, 11:53:42 PM »
Here you are moving the number 4 into r8
Code: [Select]
mov     r8,    [sum]  ; sum = 4

Here you are dividing by the number at address 4!
Code: [Select]
        idiv    dword[r8]       
 

Try:
Code: [Select]
idiv r8

Warning: untested!

Best,
Frank




Offline debs3759

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Re: Using idiv
« Reply #2 on: January 13, 2022, 11:57:41 PM »
idiv dword[r8] divides rax by the value stored at the memory address pointed to by r8. You want

Code: [Select]
idiv    r8
I also don't think using dword (32-bits) with a 64-bit register would be right, and am sure you don't need it anyway.
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Offline JohnG

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Re: Using idiv
« Reply #3 on: January 14, 2022, 12:01:52 AM »
Hi Frank,

I tried that, and I get a message   "operation size not specified"



Anyway to avoid that damn Anti-spam task ?

Offline JohnG

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Re: Using idiv
« Reply #4 on: January 14, 2022, 12:10:08 AM »
hi,

Made the other changes and program just crashes, no errors showing when compiling.

Offline Frank Kotler

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Re: Using idiv
« Reply #5 on: January 14, 2022, 01:10:18 AM »
Hi JohnG,
"other changes"? Just remove the square brackets from r8.
 
I do not get that error.

I think if you post a few more times, the  Forum will believe you are human.. Sorry 'bout that.

Best.
Frank




Offline JohnG

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Re: Using idiv
« Reply #6 on: January 14, 2022, 02:26:51 AM »
Hi all,

Yes Frank the Forum says I am human now, that could be argued by many though.

I put the x64dbg on it and it does seem to be dividing ok, some issue with the printf, which was working but not now, go figure.

Offline Frank Kotler

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Re: Using idiv
« Reply #7 on: January 14, 2022, 03:32:16 AM »
Good there is some progress with the forum, anyway.

I haven't run Windows since win98 was current. 32 bit code was similar Linux and Windows. 64 bit code not so! I can't help you much.

Code examples are always welcome!

Best,
Frank


Offline JohnG

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Re: Using idiv
« Reply #8 on: January 14, 2022, 10:06:45 PM »
Hi all,

ok, got the little test program running again.

Back to the idiv.   I see you get a quotient and a remainder  but what if you want to know the decimal part.

as in   155/4 = 38.75    what I get is  38 in rax,  3 in rdx   which is  0.75 ie 3/4


John

Offline Frank Kotler

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Re: Using idiv
« Reply #9 on: January 15, 2022, 12:33:07 AM »
Oh my!

As the name suggests, idiv is integer division. You want floating point.

In the good old days, the floating point unit, if you had one, would do this with "fdiv", Gettin the number tp/from float to ascii was the tricky part.

Nowadays, it's done with xmm or mmx - I'm not even sure which. Way above my pay grade.

Good luck!

Best,
Frank

?
Section A.4.60: DIVPD: Packed Double-Precision FP Divide
?


Offline Frank Kotler

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Re: Using idiv
« Reply #10 on: January 15, 2022, 01:37:47 AM »
This is not even close to what you want. An old DOS com file. Won't even run on modern hardware. It does show a decimal point.

Best,
Frank


Code: [Select]
;-----------------------------------------------
; converts centigrade to fahrenheit, & visa versa
;
; nasm -f bin -o myfile.com myfile.asm
;--------------------------------------

org 100h

BUFSIZ equ 8

section .data

    prompt0 db 13, 10, 7, 'Pay Attention!!!', 13, 10, '$'
    prompt1 db 13, 10, 'This humble program will convert', 13, 10
            db 'degrees centigrade to degrees Fahrenheit,', 13, 10
            db 'or Fahrenheit to centigrade (Celsius).', 13, 10
            db 'Please enter "C" or "F" to indicate the', 13, 10
            db 'temperature units in which you will INPUT', 13, 10
            db 'the value to be converted.', 13, 10, 10  , '$'
    promptc db 'Please enter degrees C to convert to F:', 13, 10, '$'
    promptf db 'Please enter degrees F to convert to C:', 13, 10, '$'
    that    db 13, 10, "That's $"
    degrees db ' degrees $'
    bye     db 13, 10, 10, "Thanks for playing!", 13, 10, '$'


    word_10 dw 10   ; multiplier for the float-to-ascii routine

                    ; for C->F, F->C routines
    float_32 dq 32.0
    float_5  dq 5.0
    float_9  dq 9.0

section .bss

    buffer resb BUFSIZ + 2     ; buffer for our input string
    result_units resb 1        ; holds 'C' or 'F'
    numbuf resb 20h            ; buffer for ftoa
    bcdbuf rest 1              ; 10 byte scratch area for ftoa
;--------------------

section .text

; cheap-a** cls - reset text mode
    mov ax, 3
    int 10h

instruct:
; tell 'em what we're gonna do, and ask for input
    mov dx, prompt1
    mov ah, 9
    int 21h

; get input - one key, no echo
    mov ah, 8
    int 21h

; force uppercase
    and al, 0DFh

; did they get it right?
    cmp al, 'C'
    je do_c2f

    cmp al, 'F'
    je do_f2c

; smack their a**, and make 'em do it over
    mov dx, prompt0
    mov ah, 9
    int 21h
    jmp short instruct

; store the letter we'll display at the end
; we also use it as a "flag" to decide f->c or c->f
; and set up appropriate prompt
do_c2f:
    mov byte [result_units], 'F'
    mov dx, promptc
    jmp short both
do_f2c:
    mov byte [result_units], 'C'
    mov dx, promptf

both:
; display the prompt, "C" or "F", asking for input
    mov ah, 9
    int 21h

; get string input, set "max size" first
    mov byte [buffer], BUFSIZ
    mov ax, 0C0Ah
    mov dx, buffer
    int 21h

; since atof expects a zero-terminated string, do so.
    mov bl, [buffer + 1]
    xor bh, bh
    mov byte [buffer + 2 + bx], 0

; convert the string to a float in st(0)
; remember that input text starts at 2 bytes into the buffer
    mov si, buffer + 2
    call atof
    jnc good_num

; if they gave us trash, give 'em trash back
; and make 'em do it over
    mov dx, prompt0
    int 21h
    mov dx, promptc
    cmp byte [result_units], 'F'
    je both
    mov dx, promptf
    jmp short both

good_num:
; okay, they did right, do the calculation they want
    cmp byte [result_units], 'C'
    jne calc_c2f

    fsub qword [float_32]
    fmul qword [float_5]
    fdiv qword [float_9]
    jmp short display

calc_c2f:
    fmul qword [float_9]
    fdiv qword [float_5]
    fadd qword [float_32]
; Whew! That was tough!

display:
; convert st(0) to ascii string at di - cx decimal places
    mov cx, 2
    mov di, numbuf
    call ftoa

; ftoa returns a zero-terminated string,
; fix it for int 21h/9 - '$'-terminate
    dec di
reterminate:
    cmp byte [di + 1], 1
    inc di
    jnc reterminate
    mov byte [di], '$'

; tell 'em what we're going to do
    mov ah, 9
    mov dx, that
    int 21h

 ; tell 'em the number
    mov dx, numbuf
    int 21h

; and the units
    mov dx, degrees
    int 21h

    mov ah, 2
    mov dl, [result_units]
    int 21h

; say Buhbye - enhancement - go again?
    mov ah, 9
    mov dx, bye
    int 21h

exit:
    mov ah,4Ch
    int 21h

;------------------------------------------------
; atof - converts ascii string to float
; expects: si pointed to zero-terminated string
; returns: float in st(0)
;          carry set if invalid digit encountered
;-------------------------------------------------
atof:
    push ax
    push bx
    push cx
    push dx
    push di
    push si

    xor bx, bx

    xor cx, cx
    dec cx                    ; set cx FFFF - no point found

    xor dx, dx
    mov byte [bcdbuf + 9], 0  ; assume positive

    mov di, si                ; save our "beginning of buffer"

.af0:
    mov al, [si]
    inc si
    cmp al, '-'
    jnz .notneg
    mov byte [bcdbuf + 9], 80h
    jmp short .af0
.notneg:
    cmp al, '.'
    jnz .notpt
    inc cx
    jmp short .af0
.notpt:
    cmp al, ' '
    jz .af0
    cmp cx, 0FFFFh
    jz .af05
    inc cx
.af05
    or al, al        ; end of string?
    jnz .af0
    dec si
    dec si
    cmp cx, 0FFFFh
    jnz .af07
    xor cx, cx
    jmp short .af1
.af07
    dec cx

.af1:
    cmp si, di
    jc .done
    mov al, [si]
    dec si
    cmp al, ' '
    jz .af1
    cmp al, '-'
    jz .af1
    cmp al, '.'
    jz .af1
    cmp al, 3Ah
    jnc .invalid
    cmp al, 30h
    jc .invalid
    and al, 0Fh
    or dh, dh
    jnz .notfirst
    or dl, al
    inc dh
    jmp short .af1
.notfirst:
    shl al, 4
    or dl, al
    mov [bcdbuf + bx], dl
    xor dx, dx
    inc bx
    jmp .af1
.invalid
    stc
    jmp short .done3

.done
    or dh, dh
    jz .padbcd
    and dl, 0Fh
    mov [bcdbuf + bx], dl
    inc bx

.padbcd
    cmp bx, byte 9
    jz .done2
    mov byte [bcdbuf + bx], 0
    inc bx
    jmp short .padbcd

.done2
    fbld [bcdbuf]
    or cx, cx
    jz .done25
.tendiv:
    fidiv word [word_10]
    loop .tendiv
.done25
    clc
.done3:

    pop si
    pop di
    pop dx
    pop cx
    pop bx
    pop ax
    ret
;--------------------------------------------


;-----------------------------------------------------
; ftoa - converts floating point to string
;
; Based on some code "sponged" from a post to clax
; From: "Jim Morrison" <astrolabe at ntplx dot net>
;
; Expects: Number to convert is on stack top - st(0)
;          di points to buffer to store string
;          cx = Decimal Places.
;-------------------------------------------------------------

ftoa:
    push ax
    push cx
    push dx
    push di
    push si

    mov dx, cx           ; save a copy of dec places
                         ; if no decimal (integer)
    jcxz .f2a2           ; skip multiply by ten loop
.f2a1:                   ; else loop to "scale" number
    fimul word [word_10]
    loop .f2a1
.f2a2:
    fbstp [bcdbuf]       ; convert to bcd and store
    mov si, bcdbuf       ; we'll pull digits from there
    mov cx, 9
.f2a3:
    lodsb                ; get a pair of digits
    mov ah, al           ; move a copy to ah
    shr ah, 4            ; shift out low nibble, keeping high
    and ax, 0F0Fh        ; mask out the digits we want
    add ax, 3030h        ; convert 'em both to ascii
    push ax
    mov al, ah           ; swap and store the other digit
    push ax
    loop .f2a3           ; until done

    cmp byte [bcdbuf+9], 0 ; sign flag at bcdbuf + 9 ?
    je .f2a6
    mov al, '-'           ; minus sign if we need it
    stosb                 ; store it at front of our string
.f2a6:

    mov cx, 18
    xor dh, dh
    inc dl
.poploop
    pop ax
    cmp al, '0'
    jnz .store
    or dh, dh
    jnz .store
    jmp short .nostore
.store
    stosb
    inc dh
.nostore
    cmp cl, dl
    jne .nopoint
    or dh, dh       ; if we haven't encountered a non-zero
    jnz .nolz
    mov al, '0'     ; put a zero before the decimal point
    stosb
.nolz
    mov al, '.'           ; decimal point
    stosb                 ; and store it
    inc dh
.nopoint
    loop .poploop

    mov al, 0
    stosb

    pop si
    pop di
    pop dx
    pop cx
    pop ax

    ret
;--------------------------------------------------


Offline JohnG

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Re: Using idiv
« Reply #11 on: January 15, 2022, 01:52:56 AM »
Hi,

yeah, I read about the xmm0/1/2 etc  registers but not sure how it all goes together. Every example I have read only do the easy divide like   90/9  so there is no remainder.

I will have a read of the example you posted here, any hints are good.

Offline JohnG

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Re: Using idiv
« Reply #12 on: January 16, 2022, 10:49:14 PM »
Hi All,

Finally got my little testing program to work with divide (floating point)

A page I found this morning may be of use to other also.

https://github.com/PoCc001/ASM-Math/blob/main/SSE/math64.asm


« Last Edit: January 16, 2022, 10:51:44 PM by JohnG »

 

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