”;
The INC Instruction
The INC instruction is used for incrementing an operand by one. It works on a single operand that can be either in a register or in memory.
Syntax
The INC instruction has the following syntax −
INC destination
The operand destination could be an 8-bit, 16-bit or 32-bit operand.
Example
INC EBX ; Increments 32-bit register INC DL ; Increments 8-bit register INC [count] ; Increments the count variable
The DEC Instruction
The DEC instruction is used for decrementing an operand by one. It works on a single operand that can be either in a register or in memory.
Syntax
The DEC instruction has the following syntax −
DEC destination
The operand destination could be an 8-bit, 16-bit or 32-bit operand.
Example
segment .data count dw 0 value db 15 segment .text inc [count] dec [value] mov ebx, count inc word [ebx] mov esi, value dec byte [esi]
The ADD and SUB Instructions
The ADD and SUB instructions are used for performing simple addition/subtraction of binary data in byte, word and doubleword size, i.e., for adding or subtracting 8-bit, 16-bit or 32-bit operands, respectively.
Syntax
The ADD and SUB instructions have the following syntax −
ADD/SUB destination, source
The ADD/SUB instruction can take place between −
- Register to register
- Memory to register
- Register to memory
- Register to constant data
- Memory to constant data
However, like other instructions, memory-to-memory operations are not possible using ADD/SUB instructions. An ADD or SUB operation sets or clears the overflow and carry flags.
Example
The following example will ask two digits from the user, store the digits in the EAX and EBX register, respectively, add the values, store the result in a memory location ”res” and finally display the result.
SYS_EXIT equ 1 SYS_READ equ 3 SYS_WRITE equ 4 STDIN equ 0 STDOUT equ 1 segment .data msg1 db "Enter a digit ", 0xA,0xD len1 equ $- msg1 msg2 db "Please enter a second digit", 0xA,0xD len2 equ $- msg2 msg3 db "The sum is: " len3 equ $- msg3 segment .bss num1 resb 2 num2 resb 2 res resb 1 section .text global _start ;must be declared for using gcc _start: ;tell linker entry point mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, msg1 mov edx, len1 int 0x80 mov eax, SYS_READ mov ebx, STDIN mov ecx, num1 mov edx, 2 int 0x80 mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, msg2 mov edx, len2 int 0x80 mov eax, SYS_READ mov ebx, STDIN mov ecx, num2 mov edx, 2 int 0x80 mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, msg3 mov edx, len3 int 0x80 ; moving the first number to eax register and second number to ebx ; and subtracting ascii ''0'' to convert it into a decimal number mov eax, [num1] sub eax, ''0'' mov ebx, [num2] sub ebx, ''0'' ; add eax and ebx add eax, ebx ; add ''0'' to to convert the sum from decimal to ASCII add eax, ''0'' ; storing the sum in memory location res mov [res], eax ; print the sum mov eax, SYS_WRITE mov ebx, STDOUT mov ecx, res mov edx, 1 int 0x80 exit: mov eax, SYS_EXIT xor ebx, ebx int 0x80
When the above code is compiled and executed, it produces the following result −
Enter a digit: 3 Please enter a second digit: 4 The sum is: 7
The program with hardcoded variables −
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
The MUL/IMUL Instruction
There are two instructions for multiplying binary data. The MUL (Multiply) instruction handles unsigned data and the IMUL (Integer Multiply) handles signed data. Both instructions affect the Carry and Overflow flag.
Syntax
The syntax for the MUL/IMUL instructions is as follows −
MUL/IMUL multiplier
Multiplicand in both cases will be in an accumulator, depending upon the size of the multiplicand and the multiplier and the generated product is also stored in two registers depending upon the size of the operands. Following section explains MUL instructions with three different cases −
| Sr.No. | Scenarios |
|---|---|
| 1 |
When two bytes are multiplied − The multiplicand is in the AL register, and the multiplier is a byte in the memory or in another register. The product is in AX. High-order 8 bits of the product is stored in AH and the low-order 8 bits are stored in AL.
|
| 2 |
When two one-word values are multiplied − The multiplicand should be in the AX register, and the multiplier is a word in memory or another register. For example, for an instruction like MUL DX, you must store the multiplier in DX and the multiplicand in AX. The resultant product is a doubleword, which will need two registers. The high-order (leftmost) portion gets stored in DX and the lower-order (rightmost) portion gets stored in AX.
|
| 3 |
When two doubleword values are multiplied − When two doubleword values are multiplied, the multiplicand should be in EAX and the multiplier is a doubleword value stored in memory or in another register. The product generated is stored in the EDX:EAX registers, i.e., the high order 32 bits gets stored in the EDX register and the low order 32-bits are stored in the EAX register.
|
Example
MOV AL, 10 MOV DL, 25 MUL DL ... MOV DL, 0FFH ; DL= -1 MOV AL, 0BEH ; AL = -66 IMUL DL
Example
The following example multiplies 3 with 2, and displays the result −
section .text global _start ;must be declared for using gcc _start: ;tell linker entry point mov al,''3'' sub al, ''0'' mov bl, ''2'' sub bl, ''0'' mul bl add al, ''0'' mov [res], al 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 section .data msg db "The result is:", 0xA,0xD len equ $- msg segment .bss res resb 1
When the above code is compiled and executed, it produces the following result −
The result is: 6
The DIV/IDIV Instructions
The division operation generates two elements – a quotient and a remainder. In case of multiplication, overflow does not occur because double-length registers are used to keep the product. However, in case of division, overflow may occur. The processor generates an interrupt if overflow occurs.
The DIV (Divide) instruction is used for unsigned data and the IDIV (Integer Divide) is used for signed data.
Syntax
The format for the DIV/IDIV instruction −
DIV/IDIV divisor
The dividend is in an accumulator. Both the instructions can work with 8-bit, 16-bit or 32-bit operands. The operation affects all six status flags. Following section explains three cases of division with different operand size −
| Sr.No. | Scenarios |
|---|---|
| 1 |
When the divisor is 1 byte − The dividend is assumed to be in the AX register (16 bits). After division, the quotient goes to the AL register and the remainder goes to the AH register.
|
| 2 |
When the divisor is 1 word − The dividend is assumed to be 32 bits long and in the DX:AX registers. The high-order 16 bits are in DX and the low-order 16 bits are in AX. After division, the 16-bit quotient goes to the AX register and the 16-bit remainder goes to the DX register.
|
| 3 |
When the divisor is doubleword − The dividend is assumed to be 64 bits long and in the EDX:EAX registers. The high-order 32 bits are in EDX and the low-order 32 bits are in EAX. After division, the 32-bit quotient goes to the EAX register and the 32-bit remainder goes to the EDX register.
|
Example
The following example divides 8 with 2. The dividend 8 is stored in the 16-bit AX register and the divisor 2 is stored in the 8-bit BL register.
section .text global _start ;must be declared for using gcc _start: ;tell linker entry point mov ax,''8'' sub ax, ''0'' mov bl, ''2'' sub bl, ''0'' div bl add ax, ''0'' mov [res], ax 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 section .data msg db "The result is:", 0xA,0xD len equ $- msg segment .bss res resb 1
When the above code is compiled and executed, it produces the following result −
The result is: 4
”;