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Assembly - Numbers
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 pnker 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 Multippcation
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 pnker 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 −
AAM − ASCII Adjust After Multippcation
AAD − ASCII Adjust Before Division
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 −
DAA − Decimal Adjust After Addition
DAS − decimal Adjust After Subtraction
There is no support for multippcation and spanision 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 pnker 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: 35801Advertisements