by Jesmin Akther | Jul 1, 2021 | Assembly Language
Procedures and Recursion used in Assembly Language:
By nature in Assembly there is an inclination towards to be big size of Assembly program. Also there is some way to keep remain in smaller in code size by subroutines or procedures. The article is concerned with the procedures and its various kind of implementation with example.
Procedures in Assembly
Assembly language programs intend to be large in size. Procedures or subroutines are very vital in assembly language which are identified by a name. Following this name, the body of the procedure is defined which accomplishes a well-defined job. End of the procedure is indicated by a return statement.
Syntax
proc_name:
procedure body
...
ret
The procedure is called from another function by using the CALL instruction. The CALL instruction should have the name of the called procedure as an argument as shown below:
CALL proc_name
The called procedure returns the control to the calling procedure by using the RET instruction.
Example: Write a very simple procedure named sum that adds the variables stored in the ECX and EDX register and returns the sum in the EAX register.
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov ecx,'4'
sub ecx, '0'
mov edx, '5'
sub edx, '0'
call sum ;call sum procedure
mov [res], 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, res
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
sum:
mov eax, ecx
add eax, edx
add eax, '0'
ret
section .data
msg db "The sum is:", 0xA,0xD
len equ $- msg
segment .bss
res resb 1
OUTPUT:
The sum is:
9
Stacks Data Structure
A stack is an array-like data structure in the memory in which data can be stored and removed from a location called the ‘top’ of the stack. The data desires to be stored is ‘pushed’ into the stack and data to be retrieved is ‘popped’ out from the stack in LIFO structure. So, the data stored first is retrieved last. Assembly language provides two instructions for stack operations: PUSH and POP. These instructions have syntaxes like:
PUSH operand
POP address/register
The memory space used in the stack segment is used for implementing stack. The registers SS and ESP (or SP) are used for implementing the stack. The top of the stack, which points to the last data item injected into the stack is pointed to by the SS: ESP register, where the SS register points to the opening of the stack segment and the SP (or ESP) gives the offset into the stack segment. The stack implementation has the following characteristics:
Only words or doublewords could be saved into the stack, not a byte. The stack grows in the reverse direction. For example, toward the lower memory address. The top of the stack points to the last item inserted in the stack; it points to the lower byte of the last word inserted. For storing the values of the registers in the stack before using them for some use; it can be done in following way:
; Save the AX and BX registers in the stack
PUSH AX
PUSH BX
; Use the registers for other purpose
MOV AX, VALUE1
MOV BX, VALUE2
...
MOV VALUE1, AX
MOV VALUE2, BX
; Restore the original values
POP BX
POP AX
Example
The following program displays the entire ASCII character set. The main program calls a procedure named display, which displays the ASCII character set.
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
call display
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
display:
mov ecx, 256
next:
push ecx
mov eax, 4
mov ebx, 1
mov ecx, achar
mov edx, 1
int 80h
pop ecx
mov dx, [achar]
cmp byte [achar], 0dh
inc byte [achar]
loop next
ret
section .data
achar db '0'
OUTPUT:
0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}
…
…
Recursion in Assembly
A recursive procedure is one that calls itself. There are two types of recursion:
In direct recursion, the procedure calls itself and in indirect recursion, the first procedure calls a second procedure, which in turn calls the first procedure. Recursion could be observed in numerous mathematical algorithms. For instance, suppose the case of calculating the factorial of a number. Factorial of a number is given by the equation:
Fact (n) = n * fact (n-1) for n > 0
For instance: factorial of 5 is 1 x 2 x 3 x 4 x 5 = 5 x factorial of 4 and this can be a good instance of showing a recursive procedure. Every recursive algorithm must have an ending condition. In the case of factorial algorithm, the end condition is reached when n is 0.
Example: The following program shows how factorial n is implemented in assembly language. To keep the program simple, we will calculate factorial 3.
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov bx, 3 ;for calculating factorial 3
call proc_fact
add ax, 30h
mov [fact], 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,fact ;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
proc_fact:
cmp bl, 1
jg do_calculation
mov ax, 1
ret
do_calculation:
dec bl
call proc_fact
inc bl
mul bl ;ax = al * bl
ret
section .data
msg db 'Factorial 3 is:',0xa
len equ $ - msg
section .bss
fact resb 1
OUTPUT:
Factorial 3 is:
6
by Jesmin Akther | Jul 1, 2021 | Assembly Language
Conditions and Loop with code in Assembly
Conditional run in assembly language is accompanied by many looping and branching instructions. Firstly we discuss here with conditions in assembly then the loops in Assembly.
Conditions in Assembly
Conditional execution in assembly language is accomplished by several looping and branching instructions. These instructions can change the flow of control in a program. Conditional execution is observed in two scenarios:
- Unconditional jump
This is performed by the JMP instruction. Conditional execution often involves a transfer of control to the address of an instruction that does not follow the currently executing instruction. Transfer of control may be forward, to execute a new set of instructions or backward, to re-execute the same steps.
2. Conditional jump
This is performed by a set of jump instructions j<condition> depending upon the condition. The conditional instructions transfer the control by breaking the sequential flow and they do it by changing the offset value in IP.
Unconditional Jump
This is performed by the JMP instruction. Conditional execution often involves a transfer of control to the address of an instruction that does not follow the currently executing instruction. Transfer of control may be forward, to execute a new set of instructions or backward, to re-execute the same steps.
Syntax
The JMP instruction provides a label name where the flow of control is transferred immediately. The syntax of the JMP instruction is:
Example
MOV AX, 00 ; Initializing AX to 0
MOV BX, 00 ; Initializing BX to 0
MOV CX, 01 ; Initializing CX to 1
L20:
ADD AX, 01 ; Increment AX
ADD BX, AX ; Add AX to BX
SHL CX, 1 ; shift left CX, this in turn doubles the CX value
JMP L20 ; repeats the statements
Conditional Jump
If some specified condition is satisfied in conditional jump, the control flow is transferred to a target instruction. There are numerous conditional jump instructions depending upon the condition and data. The syntax for the J<condition> set of instructions.
Example:
CMP AL, BL
JE EQUAL
CMP AL, BH
JE EQUAL
CMP AL, CL
JE EQUAL
NON_EQUAL: ...
EQUAL: ...
Example
The following program displays the largest of three variables. The variables are double-digit variables. The three variables num1, num2 and num3 have values 47, 22 and 31, respectively:
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov ecx, [num1]
cmp ecx, [num2]
jg check_third_num
mov ecx, [num2]
check_third_num:
cmp ecx, [num3]
jg _exit
mov ecx, [num3]
_exit:
mov [largest], ecx
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,largest
mov edx, 2
mov ebx,1 ;file descriptor (stdout)
mov eax,4 ;system call number (sys_write)
int 0x80 ;call kernel
mov eax, 1
int 80h
section .data
msg db "The largest digit is: ", 0xA,0xD
len equ $- msg
num1 dd '47'
num2 dd '22'
num3 dd '31'
segment .bss
largest resb 2
OUTPUT:
The largest digit is:
47
Loop uses in Assembly
The JMP instruction can be used for implementing loops. For example, the following code snippet can be used for executing the loop-body 10 times.
MOV CL, 10
L1:
<LOOP-BODY>
DEC CL
JNZ L1
The processor instruction set, however, includes a group of loop instructions for implementing iteration. The basic LOOP instruction has the following syntax :
LOOP label
Where, label is the target label that identifies the target instruction as in the jump instructions. The LOOP instruction assumes that the ECX register contains the loop count. When the loop instruction is executed, the ECX register is decremented and the control jumps to the target label, until the ECX register value, i.e., the counter reaches the value zero. The above code snippet could be written as:
mov ECX,10
l1:
<loop body>
loop l1
Example
The following program prints the number 1 to 9 on the screen:
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov ecx,10
mov eax, '1'
l1:
mov [num], eax
mov eax, 4
mov ebx, 1
push ecx
mov ecx, num
mov edx, 1
int 0x80
mov eax, [num]
sub eax, '0'
inc eax
add eax, '0'
pop ecx
loop l1
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .bss
num resb 1
OUTPUT
123456789:
by Jesmin Akther | Jul 1, 2021 | Assembly Language
Assembly Logical Instructions
The processor instruction set offers the instructions Boolean logic namely AND, OR, XOR, TEST, and NOT. This is tests, sets, and clears the bits according to the need of the program.
The format for these instructions:
|
Instruction
|
Format
|
| AND |
AND operand1, operand2 |
| OR |
OR operand1, operand2 |
| XOR |
XOR operand1, operand2 |
| TEST |
TEST operand1, operand2 |
The first operand in all the cases could be either in register or in memory. The second operand could be either in register/memory or an immediate or constant value. Although, memory-to-memory operations are not possible. These instructions compare or match bits of the operands and set the CF, OF, PF, SF and ZF flags.
The AND Instruction
The AND instruction is used for logical expressions bitwise AND operation. The operation returns 1, if the matching bits from both the operands are 1, otherwise it returns 0. For example:
Operand1: 0101
Operand2: 0011
——————————————–
After AND Operation -> Operand1:0001
The AND operation can be used for clearing one or more bits. For example, say the BL register contains 0011 1010. If you need to clear the high-order bits to zero, you AND it with 0FH.
AND BL, 0FH ; This sets BL to 0000 1010
Example
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov ax, 8h ;getting 8 in the ax
and ax, 1 ;and ax with 1
jz evnn
mov eax, 4 ;system call number (sys_write)
mov ebx, 1 ;file descriptor (stdout)
mov ecx, odd_msg ;message to write
mov edx, len2 ;length of message
int 0x80 ;call kernel
jmp outprog
evnn:
mov ah, 09h
mov eax, 4 ;system call number (sys_write)
mov ebx, 1 ;file descriptor (stdout)
mov ecx, even_msg ;message to write
mov edx, len1 ;length of message
int 0x80 ;call kernel
outprog:
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .data
even_msg db 'Even Number!' ;message showing even number
len1 equ $ - even_msg
odd_msg db 'Odd Number!' ;message showing odd number
len2 equ $ - odd_msg
OUTPUT
Even Number!
Change the value in the ax register with an odd digit, such as:
mov ax, 9h ; getting 9 in the ax
The program would display:
Odd Number!
Similarly to clear the entire register you can AND it with 00H.
The OR Instruction
The OR instruction is used for performing bitwise OR operation. The operations returns 1, if the matching bits from either or both operands are one. It returns 0, if both the bits are zero.
For example,
Operand1: 0101
Operand2: 0011
——————————————–
After OR -> Operand1: 0111
The OR operation can be used for setting one or more bits. For example, let us assume the AL register contains 0011 1010, you need to set the four low-order bits, you can OR it with a value 0000 1111, i.e., FH.
OR BL, 0FH ; This sets BL to 0011 1111
Example
The following example demonstrates the OR instruction. Let us store the value 5 and 3 in the AL and the BL registers, respectively, then the instruction,
OR AL, BL
should store 7 in the AL register:
section .text
global _start ;must be declared for using gcc
_start: ;tell linker entry point
mov al, 5 ;getting 5 in the al
mov bl, 3 ;getting 3 in the bl
or al, bl ;or al and bl registers, result should be 7
add al, byte '0' ;converting decimal to ascii
mov [result], al
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 0x80
outprog:
mov eax,1 ;system call number (sys_exit)
int 0x80 ;call kernel
section .bss
result resb 1
OUTPUT
7
The XOR Instruction
The XOR instruction implements from the bitwise XOR operation. The XOR operation sets the resultant bit to 1, if and only if the bits from the operands are different. If the bits from the operands are same (both 0 or both 1), the resultant bit is cleared to 0.
For example,
Operand1: 0101
Operand2: 0011
—————————————
After XOR -> Operand1: 0110
XORing an operand with itself changes the operand to 0. This is used to clear a register.
XOR EAX, EAX
The TEST Instruction
The TEST instruction works similar with the AND operation, but unlike AND instruction, it does not change the first operand. So, if we need to check whether a number in a register is even or odd, we can also do this using the TEST instruction without changing the original number.
TEST AL, 01H
JZ EVEN_NUMBER
The NOT Instruction
The NOT instruction is implementing the bitwise NOT operation. The operation reverses the bits in an operand. The operand could be either in a register or in the memory.
For example,
Operand1: 0101 0011
After NOT -> Operand1: 1010 1100
by Jesmin Akther | Jun 4, 2021 | Assembly Language
Assembly language program practice
In the previous article we have share many basic resources and other compile program with Assembly language. Do and keep practice with assembly helps to gather better knowledge on the language.
Example
section .data ;This statement for Data segment
userMsg db 'Please enter a number: ' ; This is for the user to enter a number
lenUserMsg equ $-userMsg ; This is the length of the message
dispMsg db 'You have enter: '
lenDispMsg equ $-dispMsg
section .bss ;This is Uninitialized data
num resb 5
section .text ;This Code Segment
global _start
_start: ;This is for User prompt
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h ;Read and store the user input
mov eax, 3
mov ebx, 2
mov ecx, num
mov edx, 5 ;5 bytes (numeric, 1 for sign) of that information
int 80h ;Output the message 'The enter number is: '
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h ;Output the number enter
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 5
int 80h ; Exit code
mov eax, 1
mov ebx, 0
int 80h
OUTPUT:
Please enter a number:
1234
You have enter:1234
Most assembly language instructions require operands to be processed. An operand address provides the location, where the data to be processed is stored. Some instructions do not require an operand, whereas some other instructions may require one, two, or three operands. Generally, in assembly language an instruction needs two operands, the first operand is the destination, where data in a register or memory location and the second operand is the source. Source contains either the data to be delivered that is immediate addressing or the address that means in register / memory of the data. The source data remains unchanged after the operation.
The three basic modes of addressing are:
- Register addressing
- Immediate addressing
- Memory addressing
Register Addressing:
A register which contains the operand. In this addressing mode this is the first concern. In order to depending upon the instruction, the register may be the first operand, the second operand or both.
For example,
MOV DX, TAX_RATE ; This is the Register in first operand
MOV COUNT, CX ; This is the Register in second operand
MOV EAX, EBX ; This is Both the operands are in registers
As handing out data between registers does not comprise memory, it provides fastest processing of data.
Immediate Addressing:
In an immediate operand, it has a constant value or an expression. Therefore, when an instruction with two operands uses immediate addressing, the first operand may be a register or memory location, and the second operand is an immediate constant. The first operand defines the length of the data.
For example,
BYTE_VALUE DB 150 ; This is A byte value is defined
WORD_VALUE DW 300 ; This statement A word value is defined
ADD BYTE_VALUE, 65 ; This statement An immediate operand 65 is added
MOV AX, 45H ; This statement is Immediate constant 45H is transferred to AX
Direct Memory Addressing:
This is used, when operands are specified in memory addressing mode. In direct access to main memory, usually to the data segment, it is needed. This is a way of addressing where results in slower processing of data. In order to locate the precise location of data in memory, we need the segment start address, which is typically found in the DS register and an offset value. Hence, this offset value is also called effective address. Again, In direct addressing mode, the offset value is definite directly as part of the instruction, indicated by the variable name. The assembler calculates the offset value and maintains a symbol table, which stores the offset values of all the variables used in the program. In direct memory addressing, one of the operands refers to a memory location and the other operand references a register.
For example,
ADD BYTE_VALUE, DL ; This statement is Adds the register in the memory location
MOV BX, WORD_VALUE ; This statement is Operand from the memory is added to register
Direct-Offset Addressing
This addressing mode uses the arithmetic operators to modify an address. For instance, following definitions define tables of data;
A_BYTE_TABLE DB 14, 15, 22, 45 ; This statement is Tables of bytes
A_WORD_TABLE DW 134, 345, 564, 123 ; This statement is Tables of words
The below operations access data from the tables in the memory into registers:
MOV CL, A_BYTE_TABLE [2] ; This is used to Gets the 3rd element of the BYTE_TABLE
MOV CL, A_BYTE_TABLE + 2 ; This is used, Gets the 3rd element of the BYTE_TABLE
MOV CX, A_WORD_TABLE[3] ; This is used, Gets the 4th element of the WORD_TABLE
MOV CX, A_WORD_TABLE + 3 ; This is used , Gets the 4th element of the WORD_TABLE
Indirect Memory Addressing
This addressing mode operates the computer’s capability of Segment like Offset addressing. Normally, the base registers EBX, EBP or BX, BP and the index registers DI, SI, coded within square brackets for memory references, are used for this purpose. Indirect addressing is used for variables containing several elements like, arrays. Starting address of the array is stored in, say, the EBX register.
The following code snippet shows how to access different elements of the variable.
MINE_TABLE TIMES 10 DW 0 ; This is used to Allocates 10 words (2 bytes) each initialized to 0
MOV EBX, [MY_TABLE] ; This is used to Effective Address of MINE_TABLE in EBX
MOV [EBX], 110 ; MINE _TABLE[0] = 110
ADD EBX, 2 ; EBX = EBX +2
MOV [EBX], 123 ; MINE _TABLE[1] = 123
The MOV Instruction
MOV instruction that is used for moving data from one storage space to another. The MOV instruction takes two operands.
The syntax of the MOV instruction is −
The MOV instruction may have one of the following five forms, example statements are given below:
MOV register, register
MOV register, immediate
MOV memory, immediate
MOV register, memory
MOV memory, register
Here, Both of the operands in MOV operation should be of same size so the value of source operand remains unchanged. The MOV instruction reasons ambiguity at times. For example, look at the statements:
MOV EBX, [MINE_TABLE] ; Effective Address of MINE_TABLE in EBX
MOV [EBX], 110 ; MY_TABLE[0] = 110
It is not clear. Here, either you want to move a byte equivalent or you want a word equivalent of the number 110. In such cases, it is wise to use a type specifier. Here is a table shows some of the common type of specifiers:
Type Specifier Bytes addressed
BYTE 1
WORD 2
DWORD 4
QWORD 8
TBYTE 10
Example
There is a program which illustrates some of the concepts discussed in the post. It stores a name ‘ATM SAMSUZZAMAN’ in the data section of the memory, then changes its value to another name ‘Humayon Faridi’ programmatically and displays both the names.
section .text
global _start ;This statement must be declared for linker (ld)
_start: ; This tell linker entry point
;writing the name 'ATM SAMSUZZAMAN'
mov edx,9 ;This is for message length
mov ecx, name ; This is for message to write
mov ebx,1 ; This is for a file descriptor (stdout)
mov eax,4 ; This is system call number (sys_write)
int 0x80 ; This call kernel
mov [name], dword ' Humayon Faridi ' ; Changed the name to Humayon Faridi ;writing the name ‘Humayon Faridi’
mov edx,8 ;This for message length
mov ecx,name ;This is for a message to write
mov ebx,1 ;This is the file descriptor (stdout)
mov eax,4 ;This is the system call number (sys_write)
int 0x80 ;this call kernel
mov eax,1 ;This is system call number (sys_exit)
int 0x80 ;This call kernel
section .data
name db 'ATM SAMSUZZAMAN'
When the above code is compiled and executed, it produces the following result as:
ATM SAMSUZZAMAN Humayon Faridi
by Jesmin Akther | Jun 4, 2021 | Assembly Language
Memory Arrangement or Memory Segments
A segmented memory model splits the system memory into clusters or set of autonomous segments. Each independent segments referenced by pointers located in the segment registers. This is used to contain a specific type of data. One segment is used to hold instruction codes, another segment stores the data elements, and a third segment preserves the program stack. Though there are various memory segments such as
This segment is represented by .data section and the .bss. The .data section is used to declare the memory section, where data elements are stored for the program. A section cannot be extended after the data elements are declared, and it remainders static all over the program.
This segment .bss section is also a static or not change memory section. This section comprises buffers for data to be declared later in the program. This buffer memory is zero-filled.
This is represented by .text section. This defines an area in memory that stores the instruction codes. This is also static or unchangeable or a fixed area.
This segment that is stack contains data values passed to functions and procedures within the program. In order to speed up the processor operations, the processor includes some internal memory storage locations, called registers. Registers store data elements for processing without taking to access the memory. A partial number of registers are built into the processor chip.
Processor Registers
There are ten 32-bit and six 16-bit processor registers in IA-32 architecture. The registers are grouped into three categories, for example
- General registers,
- Control registers,
- Segment registers.
The general registers are further divided into the following categorise:
- Data registers,
- Pointer registers, and
- Index registers.
- Data Registers
There are four 32-bit data registers which are used for arithmetic, logical, and other operations. These 32-bit registers can be used in three ways, such as complete 32-bit data registers: EAX, EBX, ECX, EDX. Here, Lower halves of the 32-bit registers can be used as four 16-bit data registers: AX, BX, CX and DX. Lower and higher halves of the above-mentioned four 16-bit registers can be used as eight 8-bit data registers: AH, AL, BH, BL, CH, CL, DH, and DL.
Data Registers
Some of these data registers have specific use in arithmetical operations.
AX is the primary accumulator where this is used in input/output and most arithmetic instructions. For instance , in multiplication operation, one operand is stored in EAX or AX or AL register according to the size of the operand.
BX – is known as the base register, as it could be used in indexed addressing.
CX- is known as the count register, as the ECX, CX registers store the loop count in iterative operations.
DX- is known as the data register. This register is also used in input/output operations. It is also used with AX register along with DX for multiply and divide operations involving large values.
Pointer Registers
The pointer registers are 32-bit EIP, ESP, and EBP registers and corresponding 16-bit right portions IP, SP, and BP. There are three categories of pointer registers:
Instruction Pointer (IP):
The 16-bit IP register stores the offset address of the next instruction to be executed. IP in association with the CS register gives the complete address of the current instruction in the code segment.
Stack Pointer (SP):
The 16-bit SP register provides the offset value within the program stack. SP in association with the SS register (SS:SP) refers to be current position of data or address within the program stack.
Base Pointer (BP):
The 16-bit BP register mainly helps in referencing the parameter variables passed to a subroutine. The address in SS register is combined with the offset in BP to get the location of the parameter. BP can also be combined with DI and SI as base register for special addressing.
Pointer Registers
Index Registers
The 32-bit index registers, ESI and EDI, and their 16-bit rightmost portions. SI and DI, are used for indexed addressing and sometimes used in addition and subtraction. There are two sets of index pointers:
- Source Index (SI) − It is used as source index for string operations.
- Destination Index (DI) − It is used as destination index for string operations.
Index Registers
Control Registers
The 32-bit instruction pointer register and the 32-bit flags register combined are considered as the control registers. Many instructions involve comparisons and mathematical calculations and change the status of the flags and some other conditional instructions test the value of these status flags to take the control flow to other location. The common flag bits are:
Overflow Flag (OF): This flag is indicating the overflow of a high-order bit (leftmost bit) of data after a signed arithmetic operation.
Direction Flag (DF): This flag is determining left or right direction for moving or comparing string data. When the DF value is 0, the string operation takes left-to-right direction and when the value is set to 1, the string operation takes right-to-left direction.
Interrupt Flag (IF): This flag is governs whether the external interrupts like keyboard entry, etc., are to be ignored or processed. It disables the external interrupt when the value is 0 and enables interrupts when set to 1.
Trap Flag (TF) ): This flag is allows setting the operation of the processor in single-step mode. The DEBUG program we used sets the trap flag, so we could step through the execution one instruction at a time
Sign Flag (SF) ): This flag is shows the sign of the result of an arithmetic operation. This flag is set according to the sign of a data item following the arithmetic operation. The sign is indicated by the high-order of leftmost bit. A positive result clears the value of SF to 0 and negative result sets it to 1.
Zero Flag (ZF) ): This flag is indicates the result of an arithmetic or comparison operation. A nonzero result clears the zero flag to 0, and a zero result sets it to 1.
Auxiliary Carry Flag (AF): This flag is containing the carry from bit 3 to bit 4 following an arithmetic operation; used for specialized arithmetic. The AF is set when a 1-byte arithmetic operation causes a carry from bit 3 into bit 4.
Parity Flag (PF): This flag is indicating the total number of 1-bits in the result obtained from an arithmetic operation. An even number of 1-bits clears the parity flag to 0 and an odd number of 1-bits sets the parity flag to 1.
Carry Flag (CF): This flag is containing the carry of 0 or 1 from a high-order bit (leftmost) after an arithmetic operation. It also stores the contents of last bit of a shift or rotate operation.
Segment Registers
Segments are specific areas defined in a program for containing data, code and stack. There are three main segments:
- Code Segment: This flag is containing all the instructions to be executed. A 16-bit Code Segment register or CS register stores the starting address of the code segment.
- Data Segment: This flag is contains data, constants and work areas. A 16-bit Data Segment register or DS register stores the starting address of the data segment.
- Stack Segment: This flag is containing data and return addresses of procedures or subroutines. It is implemented as a ‘stack’ data structure. The Stack Segment register or SS register stores the starting address of the stack.
Apart from the DS, CS and SS registers, there are other extra segment registers – ES (extra segment), FS and GS, which provide additional segments for storing data. These are combines the segment address in the segment register with the offset value of the location. Look at the following simple program to understand the use of registers.
Example
The use of registers in assembly programming. This program displays 7 stars on the screen with a message:
section .text
global _start ; This is must be declared for linker (gcc)
_start: ; This tell linker entry point
mov edx,len ; This is a message length
mov ecx,msg ; This is a message to write
mov ebx,1 ; This is a file descriptor (stdout)
mov eax,4 ; This is system call number (sys_write)
int 0x80 ; This is call kernel
mov edx,7 ; This is message length
mov ecx,s2 ; This is message to write
mov ebx,1 ; This is a file descriptor (stdout)
mov eax,4 ; This is system call number (sys_write)
int 0x80 ; This is call kernel
mov eax,1 ; This is system call number (sys_exit)
int 0x80 ; This is call kernel
section .data
msg db 'Displaying 7 stars',0xa ;a message
len equ $ - msg ;length of message
s2 times 7 db '*'
OUTPUT
Displaying 7 stars
*******
by Jesmin Akther | May 31, 2021 | Assembly Language
Environment Setup
Assembly language is dependent upon the instruction set and the architecture of the processor. There are many good assembler programs, for example
- NASM: It is an operating system independent assembler. One of the two widely used Linux assemblers and the other GNU
- The GNU assembler (GAS): The syntax differs significantly in many ways from
- MASM (Microsoft Assembler): MASM syntax is a standard. So, almost always understood by other x86 assemblers TASM, CHASM, A386, etc. The syntax has some significant defects that makes coding error hence many of them are rectified in NASM.
- Borland Turbo Assembler (TASM)
Installing NASM
It could be used on both Linux and Windows, can download from various web sources. All are well documented. While installing Linux, if “Development Tools” is chacked, NASM installed along with the Linux operating system. For checking have NASM installed, take the following steps
- Open a Linux terminal.
- Type where is nasm and press ENTER.
We use a online compiler of assymbly language in this blog. Go to this link – https://www.jdoodle.com/compile-assembler-nasm-online/
Basic of NASM assembler
Character Set: Letters a..z; A..Z; ()
Digits: 0.9
Special Chars: ? _ @ $ . ~
- NASM is case-sensitive with respect to labels and variables
- It is not case-sensitive with respect to keywords, mnemonics, register names, directives, etc.
- Special Characters.
Write a basic assembly program
Generally, an assembly program can be divided into three sections, such as
- The data section,
- The bss section, and
- The text section.
The data Section
The section is used for declaring data or constants which are not modify at runtime. Various constant values, file names, or buffer size, etc. are declare in this section.
The syntax for declaring data section is “section.data”
The bss Section
The section is used for declaring variables. The syntax for declaring bss section is “section.bss”
The text section
This section must be begun with the declaration global _start, which tells the kernel where the program execution begins. The section is used for care the actual code.
The syntax for declaring text section is
section.text
global _start
_start:
Comments
AL comment begins with a semicolon (;). It may contain any printable character including blank. It can appear on a line by itself, like below
; This program displays a message on screen
or, on the same line along with an instruction, like
add eax, ebx ; this statement state as adds ebx to eax
Assembly Language Statements
Assembly language programs consist of three types of statements, for example
- Executable instructions or instructions:
The executable instructions or simply instructions tell the processor what to do. Each instruction consists of an operation code (opcode). Each executable instruction generates one machine language instruction.
- Assembler directives or pseudo-ops
The assembler directives or pseudo-ops tell the assembler about the various aspects of the assembly process. These are non-executable and do not generate machine language instructions.
- macros
In AL macros are basically a text substitution mechanism.
Syntax of Assembly Language(AL) Statements
Assembly language statements are entered one statement per line. Each statement follows the following format:
[label] mnemonic [operands] [;comment]
A basic instruction has two parts, the first one is the name of the instruction (or the mnemonic), which is to be executed, and the second are the operands or the parameters of the command. Here the fields in the square brackets are optional.
Following are some examples of typical assembly language statements :
INC COUNT ; this statement state as Increment the memory variable COUNT
MOV TOTAL, 48 ; this statement state as Transfer the value 48 in the ; memory variable TOTAL
ADD AH, BH ; this statement state as Add the content of the BH register into the AH register
AND MASK1, 128 ; this statement state as Perform AND operation on the variable MASK1 and 128
ADD MARKS, 10 ; this statement state as Add 10 to the variable MARKS
MOV AL, 10 ; this statement state as Transfer the value 10 to the AL register
The Hello World Program in Assembly Language
The following assembly language code displays the string ‘Hello World’ on the screen −
section .text
global _start ;This is must be declared for linker (ld)
_start: ; this statement tells linker entry point
mov edx,len ; this state as message length
mov ecx,msg ; this statement state as message to write
mov ebx,1 ; this statement state as file descriptor (stdout)
mov eax,4 ; this statement state as system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ; this statement state as system call number (sys_exit)
int 0x80 ;call kernel
section .data
msg db 'Hello, world!', 0xa ;string to be printed
len equ $ - msg ;length of the string
OUTPUT:
Hello, world!
Now Compiling and Linking an AL Program in NASM. Make sure you have set the path of nasm and ld binaries in your PATH environment variable. Now, take the following steps for compiling and linking the above program:
- Type the above code using a text editor and save it as hello.asm.
- Make sure that you are in the same directory as where you saved hello.asm.
- To assemble the program, type nasm -f elf hello.asm
- If there is any error, you will be prompted about that at this stage. Otherwise, an object file of your program named hello.o will be created.
- To link the object file and create an executable file named hello, type ld -m elf_i386 -s -o hello hello.o
- Execute the program by typing ./hello
- If you have done everything correctly, it will display ‘Hello, world!’ on the screen.
if replace the section keyword with segment by follow code:
segment .text
global _start ;This is must be declared for linker (ld)
_start: ; this statement tells linker entry point
mov edx,len ; this state as message length
mov ecx,msg ; this statement state as message to write
mov ebx,1 ; this statement state as file descriptor (stdout)
mov eax,4 ; this statement state as system call number (sys_write)
int 0x80 ;call kernel
mov eax,1 ; this statement state as system call number (sys_exit)
int 0x80 ;call kernel
segment .data
msg db 'Hello, world!', 0xa ;string to be printed
len equ $ - msg ;length of the string
OUTPUT
Hello, world!
by Jesmin Akther | May 31, 2021 | Assembly Language
What is Assembly Language?
Assembly Language (AL) is one line of code translates to one machine instruction. Every computer has a microprocessor that achieves the computer’s arithmetical, logical, and control actions. ALs are NOT machine-independent that is each different machine or processor has a different machine languages. Any particular machine can have more than one assembly language. This is a low-level language that represents various instructions in symbolic code and a clearer form. A processor understands only machine language instructions, which are strings of 1’s and 0’s. Each set of processors has own instructions set for handling numerous operations such as getting input from keyboard, displaying information on screen and performing many works where each set of instructions are named ‘machine language instructions’.
Assembly language makes one aware of:
- How OS, processor, and BIOS programs interface mutually;
- How data is represented in memory with other devices;
- How the processor accesses and executes instruction;
- How each instruction access to process data;
- How a program accesses external device
Assembly Language Advantages:
- The language requires less memory and execution time;
- It allows hardware-specific complex jobs in an easier way;
- It is suitable for time-critical trades;
- It is most suitable for writing interrupt service routines along other memory resident programs.
A computer Hardware Basic Features
The main internal hardware of a PC consists of processor, memory, and registers. Registers are processor components where data and address are hold. To execute a program, the system copies it from the external device into the internal memory. The processor executes the program instructions.
The fundamental unit of computer storage is a bit; it could be ON (1) or OFF (0) and a group of 8 bits and makes a byte on most of the modern computers.
The parity bit is used to make the number of bits in a byte odd. If the parity is even, the system assumes that there had been a parity error (though rare), which might have been caused due to hardware fault or electrical disturbance.
The processor supports data sizes:
- Word: Contain a 2-byte data item
- Doubleword: Consist of a 4-byte (32 bit) data item
- Quadword: Contain an 8-byte (64 bit) data item
- Paragraph: Have a 16-byte (128 bit) area
- Kilobyte: Have 1024 bytes
- Megabyte: Have 1,048,576 bytes
Binary Number System
Every number system uses positional notation. Each position is power of the base, which is 2 for binary number system, and these powers begin at 0 and increase by 1.
The positional values for an 8-bit binary number, with all bits are set ‘ON’
| Bit value |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
| Position value as a power of base 2 |
128 |
64 |
32 |
16 |
8 |
4 |
2 |
1 |
| Bit number |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
The value of a binary number is based on the presence of 1 bits and their positional value. So, the value of a given binary number is:
1 + 2 + 4 + 8 +16 + 32 + 64 + 128 = 255
which is same as 28 – 1.
Hexadecimal Number System
Hexadecimal number system uses base 16 range from 0 to 15 each digit. The letters A through F is represented corresponding to decimal values 10 through 15.
| Decimal number |
Binary representation |
Hexadecimal representation |
| 0 |
0 |
0 |
| 1 |
1 |
1 |
| 2 |
10 |
2 |
| 3 |
11 |
3 |
| 4 |
100 |
4 |
| 5 |
101 |
5 |
| 6 |
110 |
6 |
| 7 |
111 |
7 |
| 8 |
1000 |
8 |
| 9 |
1001 |
9 |
| 10 |
1010 |
A |
| 11 |
1011 |
B |
| 12 |
1100 |
C |
| 13 |
1101 |
D |
| 14 |
1110 |
E |
| 15 |
1111 |
F |
To convert a binary number to its hexadecimal equivalent, break it into groups of 4 consecutive groups each, starting from the right, and write those groups over the corresponding digits of the hexadecimal number.
Example − Binary number 1000 1100 1101 0001 is equivalent to hexadecimal – 8CD1
To convert a hexadecimal number to binary, just write each hexadecimal digit into its 4-digit binary equivalent.
Example − Hexadecimal number FCD7 is equivalent to binary – 1111 1100 1101 0111
Binary Arithmetic
The following table illustrates four simple rules for binary addition −
| (i) |
(ii) |
(iii) |
(iv) |
|
|
|
1 |
| 0 |
1 |
1 |
1 |
| +0 |
+0 |
+1 |
+1 |
| =0 |
=1 |
=10 |
=11 |
Rules (iii) and (iv) show a carry of a 1-bit into the next left position.
Example
| Decimal |
Binary |
| 60 |
00111100 |
| +42 |
00101010 |
| 102 |
01100110 |
A negative binary value is expressed in two’s complement notation. According to this rule, to convert a binary number to its negative value is to reverse its bit values and add 1.
Example
| Number 53 |
00110101 |
| Reverse the bits |
11001010 |
| Add 1 |
00000001 |
| Number -53 |
11001011 |
To subtract one value from another, convert the number being subtracted to two’s complement format and add the numbers.
Example
Subtract 22 from 33
| Number 33 |
00100001 |
| Number 22 |
00010110 |
| Reverse the bits of 42 |
11101001 |
| Add 1 |
00000001 |
| Number -22 |
00010110 |
| 33 – 22 = 11 |
00001011 |
Overflow of the last 1 bit is lost.
Addressing Data in Memory
The processor which the controls process of execution of instructions is denoted as the fetch-decode-execute cycle or the execution cycle. It consists of three continuous steps:
- Fetching the instruction from memory
- Decoding or identifying the instruction
- Executing the instruction
The processor may access one or more bytes of memory at a time. Let us consider a hexadecimal number 0225H. This number will require two bytes of memory. The high-order byte or most significant byte is 02 and the low-order byte is 25. The processor stores data in reverse-byte sequence, i.e., a low-order byte is stored in a low memory address and a high-order byte in high memory address. So, if the processor brings the value 0225H from register to memory, it will transfer 25 first to the lower memory address and 02 to the next memory address.
There are two kinds of memory addresses:
- Absolute address: It is a direct reference of specific location.
- Segment address (or offset) : It is a starting address of a memory segment with the offset value.
