Assembly - Numbers
Numerical data is generally represented in binary system. Arithmetic instructions operate on binary data. When numbers are displayed on screen or entered from keyboard, they are in ASCII form.
So far, we have converted this input data in ASCII form to binary for arithmetic calculations and converted the result back to binary. The following code shows this −
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov eax,'3'
sub eax, '0'
mov ebx, '4'
sub ebx, '0'
add eax, ebx
add eax, '0'
mov [sum], eax
mov ecx,msg
mov edx, len
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov ecx,sum
mov edx, 1
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db "The sum is:", 0xA,0xD
len equ $ - msg
segment .bss
sum resb 1
When the above code is compiled and executed, it produces the following result −
The sum is:
7
Such conversions, however, have an overhead, and assembly language programming allows processing numbers in a more efficient way, in the binary form. Decimal numbers can be represented in two forms −
- ASCII form
- BCD or Binary Coded Decimal form
ASCII Representation
In ASCII representation, decimal numbers are stored as string of ASCII characters. For example, the decimal value 1234 is stored as −
31 32 33 34H
Where, 31H is ASCII value for 1, 32H is ASCII value for 2, and so on. There are four instructions for processing numbers in ASCII representation −
AAA − ASCII Adjust After Addition
AAS − ASCII Adjust After Subtraction
AAM − ASCII Adjust After Multiplication
AAD − ASCII Adjust Before Division
These instructions do not take any operands and assume the required operand to be in the AL register.
The following example uses the AAS instruction to demonstrate the concept −
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
sub ah, ah
mov al, '9'
sub al, '3'
aas
or al, 30h
mov [res], ax
mov edx,len ;message length
mov ecx,msg ;message to write
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov edx,1 ;message length
mov ecx,res ;message to write
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db 'The Result is:',0xa
len equ $ - msg
section .bss
res resb 1
When the above code is compiled and executed, it produces the following result −
The Result is:
6
BCD Representation
There are two types of BCD representation −
- Unpacked BCD representation
- Packed BCD representation
In unpacked BCD representation, each byte stores the binary equivalent of a decimal digit. For example, the number 1234 is stored as −
01 02 03 04H
There are two instructions for processing these numbers −
The four ASCII adjust instructions, AAA, AAS, AAM, and AAD, can also be used with unpacked BCD representation. In packed BCD representation, each digit is stored using four bits. Two decimal digits are packed into a byte. For example, the number 1234 is stored as −
12 34H
There are two instructions for processing these numbers −
There is no support for multiplication and division in packed BCD representation.
Example
The following program adds up two 5-digit decimal numbers and displays the sum. It uses the above concepts −
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov esi, 4 ;pointing to the rightmost digit
mov ecx, 5 ;num of digits
clc
add_loop:
mov al, [num1 + esi]
adc al, [num2 + esi]
aaa
pushf
or al, 30h
popf
mov [sum + esi], al
dec esi
loop add_loop
mov edx,len ;message length
mov ecx,msg ;message to write
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov edx,5 ;message length
mov ecx,sum ;message to write
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db 'The Sum is:',0xa
len equ $ - msg
num1 db '12345'
num2 db '23456'
sum db ' '
When the above code is compiled and executed, it produces the following result −
The Sum is:
35801
