Assembly - Numbers

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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 the following 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:

  • AAM - ASCII Adjust After Multiplication

  • 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 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


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