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。

語法:

INC指令的語法如下:

INC destination

操作數目標可能是8位,16位或32位操作數。

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。

語法:

DEC指令的語法如下:

DEC destination

操作數目標可以是8位,16位或32位操作數。

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]

ADD和SUB指令

ADD和SUB指令用於執行二進制數據字節,字和雙字的大小,即簡單的加法/減法,8位,16位或32位操作數分別相加或相減。

語法:

ADD和SUB指令的語法如下:

ADD/SUB destination, source

ADD/ SUB指令之間可能發生:

  • 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

然而,像其他指令,內存到內存的操作是不可能使用ADD/ SUB指令。 ADD或SUB操作設置或清除溢出和進位標誌。

例子:

下面的例子會從用戶要求的兩個數字,分別在EAX和EBX寄存器存儲的數字,增加值,並將結果存儲在一個內存位置'清晰度',並最終顯示結果。

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 yiibai
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

上面的代碼編譯和執行時,它會產生以下結果:

Enter a digit:
3
Please enter a second digit:
4
The sum is:
7

該程序用硬編碼的變量:

section .text global _start ;must be declared for using gcc
_start: ;tell linker entry yiibai
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

上面的代碼編譯和執行時,它會產生以下結果:

The sum is:
7

MUL/ IMUL指令

有兩個指令乘以二進制數據。 MUL(乘)指令處理無符號數據和IMUL(整數乘法)處理有符號數據。這兩個指令影響進位和溢出標誌。

語法:

MUL/ IMUL指令,語法如下:

MUL/IMUL multiplier

在這兩種情況被乘數是在累加器中,根據被乘數和乘數的大小,所產生的產物也被存儲在操作數,大小取決於兩個寄存器。下面一節的解釋MULL有三種不同的情況指令:

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

8

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.

16

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.

32

例子:

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

例子:

下面的示例與2乘以3,並顯示結果:

section .text global _start ;must be declared for using gcc
_start: ;tell linker entry yiibai

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

上面的代碼編譯和執行時,它會產生以下結果:

The result is:
6

DIV/IDIV 指令

除法運算產生兩個元素 - 一個商和餘數。在乘法運算的情況下,不會發生溢出,因爲雙倍長度的寄存器是用來保持產生。然而,在除法的情況下,可能會發生溢出。處理器產生一箇中斷,如果發生溢出。

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

被除數是在累加器。兩個指令可以處理8位,16位或32位操作數。該操作會影響所有的6個狀態標誌。以下部分說明了三個例子的劃分有不同的操作數大小:

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.

DIV

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.

DIV

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.

DIV

例子:

下面的例子8除於2。8被存儲在16位寄存器EAX和除數2被存儲在8位BL寄存器。

section .text global _start ;must be declared for using gcc
_start: ;tell linker entry yiibai
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

上面的代碼編譯和執行時,它會產生以下結果:

The result is:
4

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