Assembly 算術指令

INC指令

INC指令是一個用於操作數遞增。它可以在一個單一的操作數，可以是在一個寄存器或內存。

INC指令的語法如下：

INC destination

例子:

INC EBX ; Increments 32-bit register INC DL ; Increments 8-bit register INC [count] ; Increments the count variable

DEC指令

DEC指令用於由一個操作數遞減。它可以在一個單一的操作數，可以是在一個寄存器或內存。

DEC指令的語法如下：

DEC destination

例子:

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]

• 寄存器到寄存器

• 內存到寄存器

• 寄存器到內存

• 寄存器到常量數據

• 內存到常量數據

例子:

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``

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, [number1] sub eax, '0' mov ebx, [number2] 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

Enter a digit:
3
4
The sum is:
7

section .text global _start ;must be declared for using gcc
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

The sum is:
7

MUL/ IMUL指令

語法:

MUL/ IMUL指令，語法如下：

MUL/IMUL multiplier

SN

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 double word, 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.

例子:

MOV AL, 10 MOV DL, 25 MUL DL ... MOV DL, 0FFH ; DL= -1 MOV AL, 0BEH ; AL = -66 IMUL DL

例子:

section .text global _start ;must be declared for using gcc

``````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

The result is:
6

DIV/IDIV 指令

DIV（除）指令或無符號數據和IDIV（整數除法）用於有符號數據。

語法:

DIV / IDIV指令的格式爲：

DIV/IDIV divisor

SN

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.

例子:

section .text global _start ;must be declared for using gcc
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

The result is:
4

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