How divided c number in rept ?

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Roman

Joined: 21 Apr 2012
Posts: 2019

Roman 25 Jan 2025, 13:41

Fasmw 1.73

Code:

rept 7 n:0,c:1000000000 { 
                     mov ebx,c
                     c equ (c/10)
                     mul ebx }

IDA Pro show:
mov ebx,1000000000
mul ebx
mov ebx,1000000001
mul ebx
mov ebx,1000000002
mul ebx

I want get this:

Code:

mov ebx,1000000000
mul ebx
mov ebx,100000000
mul ebx
mov ebx,10000000
mul ebx

25 Jan 2025, 13:41

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 20759
Location: In your JS exploiting you and your system

revolution 25 Jan 2025, 14:10

You have to use the 'c' variable outside of the rept.

Code:

c equ 1000000000
rept 7 n:0 {
        dd c
        c equ (c/10)
}

Which allows you to use the native aritmetic:

Code:

c = 1000000000
rept 7 n:0 {
        dd c
        c = c/10
}

25 Jan 2025, 14:10

Roman

Joined: 21 Apr 2012
Posts: 2019

Roman 25 Jan 2025, 14:20

Thanks.
Work fine.

25 Jan 2025, 14:20

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 20759
Location: In your JS exploiting you and your system

revolution 25 Jan 2025, 14:40

You also don't need 'n':

Code:

c = 1000000000
rept 7 {
        dd c
        c = c/10
}

25 Jan 2025, 14:40

revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 20759
Location: In your JS exploiting you and your system

revolution 25 Jan 2025, 14:59

To illustrate what happens there, try to predict what this code will display:

Code:

rept 1 c:10 { c equ (c/10) }
display 10 + '0'

It will display '1':

Code:

flat assembler  version 1.73.31  (16384 kilobytes memory)
1
1 passes, 0 bytes.

Because the string '10' has been equated to '(10/10)' in the first line, since all strings matching 'c' are replaced with '10'.

The same code with comments explaining:

Code:

rept 1 c:10 { c equ (c/10) }    ; 10 equ (10/10)
display 10 + '0'                ; display (10/10) + '0'

25 Jan 2025, 14:59

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