Division in assembly. Characters are stored in ASCII?
section .text global main main: mov esi, 10 ; ESI holds current number mov eax, 0 ; EAX holds sum of numbers push 10 ; Line feed for end of line (Stack is first in, last out -> Line feed will be last char) mov edi, 10 ; Divisor of 10 for seperating digits of number (used in divisionLoop) ; Sum numbers 1 through 10 sumLoop: add eax, esi ; Add number to sum dec esi ; Next number jnz sumLoop ; Loop until 0 ; Seperate eax into its digits, by dividing by 10 multiple times, ; where in each division the remainder will be a single digit ; and the quotient will be the remaining digits used as dividend in next loop run divisionLoop: mov edx, 0 ; Make sure edx is empty, as it is used as upper half of dividend div edi ; Divide eax by edi (= 10) => quotient is in eax (= Rest of digits for next loop), remainder in edx (= Single digit) add edx, 48 ; Make edx (digit) a char representing its value by adding '0' to it push edx ; Push char to stack for usage in print later cmp eax, 0 ; Loop until quotient is 0 (=> no more digits left) jne divisionLoop ; Print digits from Stack one by one printLoop: mov eax, 4 mov ebx, 1 mov ecx, esp ; Print top of stack (esp always points to top of stack) mov edx, 1 ; Length of 1 byte (= 1 char) int 80h pop esi ; Remove top of stack cmp esi, 10 ; Loop until Line feed is reached jne printLoop exit: mov eax, 1 mov ebx, 0 ; Exit code 0 int 80hHello,
Here is a program in assembler that adds the numbers from 1 to 10.
Divides 55 by 10 and then stores the remainders of the division in a stack.
push….
In the end everything is output to the console via the standard channel.
If I divide 55 by 10, the quotient is 5, which is in register eax, and the remainder is also 5, which is in register edx. Before I push the remainder onto the stack, the number is added to 48. Result bit 53 is the char character 5 in ASCII code.
Does this mean that the results of integer division in assembly code are always stored in ASCII characters?
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No, of course it doesn’t mean that. ASCII is not other than a character encoding, i.e. a match which “number” carries a character.
Fortunately, the numbers successively begin at 0 in the coding, so that the addition of the value from 48 to the numerical value corresponds to the character of the digit.
“No, of course it doesn’t mean that. ASCII is not other than a character encoding, i.e. a match which “number” carries a sign.”
=> Ok, Understood!
What I do not understand is the following divisions loop:
DivisionLoop:
mov edx, 0
div edi
Add edx, 48
press release
…
=> I share as written 55 by 10, resulting in a quotient of 5, which is in the register eax, and I have a sum of 5, which is in the regsiter edx.
Now here is with add edx, 48
48 added to the contents of the register edx. Ergbit 53 in ASCII corresponds to the sign 5.
=> D. h. in the register edx is originally before the addititon +48 the value 5
and this value 5 is coded in the assembler code in ASCII?
If you speak from the source, then yes, there are number liters of course ASCII coded, otherwise they would not be displayed as such.
In the finished Binary, of course, this looks different.
Or, in other words, you wrote to me that if I want to use the standart canal for output, then I can hand it over something he can do with.
As a char or a string in Assembly
How do I know that I can give the standard output channel in Assembyl just this and no number?
This is simply because the standard output channel is a console and numbers in most codings are coded as in ASCII.
If I were to redirect stdout into a file, then if you write ASCII to STDOUT, it would be just encoded ASCII, using UTF-8, then UTF-8.
The terminal/terminal emulator reads virtually your stdout and interprets it according to definition.
do Binary?