Assembly 算术指令 - Assembly汇编
INC指令
INC指令是一个用于操作数递增。它可以在一个单一的操作数，可以是在一个寄存器或内存。
语法:
INC指令的语法如下：
INC destination
操作数目标可能是8位，16位或32位操作数。
例子:
INC EBX ; Increments 32-bit register
INC DL ; Increments 8-bit register
INC [count] ; Increments the count variable
DEC指令
DEC指令用于由一个操作数递减。它可以在一个单一的操作数，可以是在一个寄存器或内存。
语法:
DEC指令的语法如下：
DEC destination
操作数目标可以是8位，16位或32位操作数。
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指令用于执行二进制数据字节，字和双字的大小，即简单的加法/减法，8位，16位或32位操作数分别相加或相减。
语法:
ADD和SUB指令的语法如下：
ADD/SUB destination, source
ADD/ SUB指令之间可能发生：
寄存器到寄存器
内存到寄存器
寄存器到内存
寄存器到常量数据
内存到常量数据
然而，像其他指令，内存到内存的操作是不可能使用ADD/ SUB指令。 ADD或SUB操作设置或清除溢出和进位标志。
例子:
下面的例子会从用户要求的两个数字，分别在EAX和EBX寄存器存储的数字，增加值，并将结果存储在一个内存位置'清晰度'，并最终显示结果。
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
上面的代码编译和执行时，它会产生以下结果：
Enter a digit:
3
Please enter a second digit:
4
The sum is:
7
该程序用硬编码的变量：
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
上面的代码编译和执行时，它会产生以下结果：
The sum is:
7
MUL/ IMUL指令
有两个指令乘以二进制数据。 MUL（乘）指令处理无符号数据和IMUL（整数乘法）处理有符号数据。这两个指令影响进位和溢出标志。
语法:
MUL/ IMUL指令，语法如下：
MUL/IMUL multiplier
在这两种情况被乘数是在累加器中，根据被乘数和乘数的大小，所产生的产物也被存储在操作数，大小取决于两个寄存器。下面一节的解释MULL有三种不同的情况指令：
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
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.
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:
下面的示例与2乘以3，并显示结果：
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
上面的代码编译和执行时，它会产生以下结果：
The result is:
6
DIV/IDIV 指令
除法运算产生两个元素 - 一个商和余数。在乘法运算的情况下，不会发生溢出，因为双倍长度的寄存器是用来保持产生。然而，在除法的情况下，可能会发生溢出。处理器产生一个中断，如果发生溢出。
DIV（除）指令或无符号数据和IDIV（整数除法）用于有符号数据。
语法:
DIV / IDIV指令的格式为：
DIV/IDIV divisor
被除数是在累加器。两个指令可以处理8位，16位或32位操作数。该操作会影响所有的6个状态标志。以下部分说明了三个例子的划分有不同的操作数大小：
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.
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.
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:
下面的例子8除于2。8被存储在16位寄存器EAX和除数2被存储在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
上面的代码编译和执行时，它会产生以下结果：
The result is:
4