Assembly 算術指令 - tw511教學網

INC指令

INC指令是一個用於運算元遞增。它可以在一個單一的運算元，可以是在一個暫存器或記憶體。

語法:

INC指令的語法如下：

INC destination

運算元目標可能是8位元，16位元或32位元運算元。

例子:

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

DEC指令

DEC指令用於由一個運算元遞減。它可以在一個單一的運算元，可以是在一個暫存器或記憶體。

語法:

DEC指令的語法如下：

DEC destination

運算元目標可以是8位元，16位元或32位元運算元。

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指令用於執行二進位制資料位元組，字和雙字的大小，即簡單的加法/減法，8位元，16位元或32位元運算元分別相加或相減。

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：

暫存器到暫存器
記憶體到暫存器
暫存器到記憶體
暫存器到常數資料
記憶體到常數資料

然而，像其他指令，記憶體到記憶體的操作是不可能使用ADD/ SUB指令。 ADD或SUB操作設定或清除溢位和進位標誌。

例子:

下面的例子會從使用者要求的兩個數位，分別在EAX和EBX暫存器儲存的數位，增加值，並將結果儲存在一個記憶體位置'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

上面的程式碼編譯和執行時，它會產生以下結果：

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

該程式用寫死的變數：

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

上面的程式碼編譯和執行時，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位制資料。 MUL（乘）指令處理無符號資料和IMUL（整數乘法）處理有符號資料。這兩個指令影響進位和溢位標誌。

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，所產生的產物也被儲存在運算元，大小取決於兩個暫存器。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
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.

Scenarios

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 bit Source Registers in MUL instruction

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 bit Source Registers in MUL instruction

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 bit Source Registers in MUL instruction

例子:

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

例子:

下面的範例與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

上面的程式碼編譯和執行時，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，不會發生溢位，因為雙倍長度的暫存器是用來保持產生。然而，在除法的情況下，可能會發生溢位。處理器產生一個中斷，如果發生溢位。

DIV（除）指令或無符號資料和IDIV（整數除法）用於有符號資料。

語法:

DIV / IDIV指令的格式為：

DIV/IDIV	divisor

被除數是在累加器。兩個指令可以處理8位元，16位元或32位元運算元。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運算元大小：

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.
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.

Scenarios

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 instruction

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 instruction

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 instruction

例子:

下面的例子8除於2。8被儲存在16位元暫存器EAX和除數2被儲存在8位元BL暫存器。

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

上面的程式碼編譯和執行時，它會產生以下結果：

The result is:
4