Converting any number into ASCII

[Stormbreaker]

Hello,

I want to convert a number into ASCII, it's easy with a one-digit-number: Just have to add 30h, but what about two-digit and so on?

[Tedd]

Divide by 10 :wink

When you do a DIV you get the answer (in eax) and the remainder (in edx).
So what you do is divide the number by 10, and use the value in eax as the last (rightmost) digit.
Then take the remainder (if it's not zero yet) and div by 10 again, to get the next digit.
Keep doing that until the remainder is zero :wink

[Vortex]

You would like to have a look at the dwtoa function from masm32.lib :

dwtoa

dwtoa proc public uses esi edi dwValue:DWORD, lpBuffer:PTR BYTE

Note that the parameter lpBuffer is an address of DWORD size.

Description
dwtoa convert a DWORD value to an ascii string.

Parameters
   1. dwValue The DWORD value to convert.

   2. lpBuffer The address of the buffer to put the converted DWORD into.

Return Value
There is no return value.

Comments
The buffer for the converted value should be large enough to hold the string including the terminating zero

[kermit]

Divide by 10 :wink

When you do a DIV you get the answer (in eax) and the remainder (in edx).
So what you do is divide the number by 10, and use the value in eax as the last (rightmost) digit.
Then take the remainder (if it's not zero yet) and div by 10 again, to get the next digit.
Keep doing that until the remainder is zero :wink

i did an example of this method for 16bit masm, just take a look at the last code example on the page:
http://www.osap.de/asm_tut/asmtut2f.html
it prints out the value thats stored in the ax register...
( page is in german but i hope the code is readable anyway  :P

[PBrennick]

kermit,
That is nice, but I think Tedd's description of using a 32bit method is better suited for today's type of code.

Paul

[zooba]

Divide by 10 :wink

When you do a DIV you get the answer (in eax) and the remainder (in edx).
So what you do is divide the number by 10, and use the value in eax as the last (rightmost) digit.
Then take the remainder (if it's not zero yet) and div by 10 again, to get the next digit.
Keep doing that until the remainder is zero :wink

You've got some of your eax's an edx's around the wrong way. :wink

So what you do is divide the number by 10, and use the remainder in edx as the last (rightmost) digit.
Then take the answer (eax) (if it's not zero yet) and div by 10 again, to get the next digit.
Keep doing that until the answer (eax) is zero

Cheers,

Zooba :U

[Tedd]

:lol Good point!

(No no, I meant to do it like that! It was an.. umm... exercise for the reader! :green2)