汇编-算术指令 - 芒果文档

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📜 汇编-算术指令

📅 最后修改于: 2020-11-05 05:06:20 🧑 作者: Mango

INC指令

INC指令用于将操作数加1。它对可以在寄存器或内存中的单个操作数起作用。

句法

INC指令具有以下语法-

INC destination

操作数目的地可以是8位，16位或32位操作数。

例

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

DEC指令

DEC指令用于将操作数减1。它对可以在寄存器或内存中的单个操作数起作用。

句法

DEC指令具有以下语法-

DEC destination

操作数目的地可以是8位，16位或32位操作数。

例

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指令用于对字节，字和双字大小的二进制数据进行简单的加/减，即分别用于添加或减去8位，16位或32位操作数。

句法

ADD和SUB指令具有以下语法-

ADD/SUB    destination, source

ADD / SUB指令可以发生在-

注册注册
内存注册
注册到内存
注册到常量数据
存储恒定数据

但是，像其他指令一样，使用ADD / SUB指令也无法进行存储器到存储器的操作。 ADD或SUB操作设置或清除溢出和进位标志。

例

下面的示例将要求用户输入两位数字，分别将这些数字存储在EAX和EBX寄存器中，将这些值相加，将结果存储在“ res ”存储位置中，最后显示结果。

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 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

编译并执行上述代码后，将产生以下结果-

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

具有硬编码变量的程序-

现场演示

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

编译并执行上述代码后，将产生以下结果-

The sum is:
7

MUL / IMUL指令

有两条指令用于将二进制数据相乘。 MUL(乘法)指令处理未签名的数据，而IMUL(整数乘法)则处理签名的数据。两条指令都影响进位和溢出标志。

句法

MUL / IMUL指令的语法如下-

MUL/IMUL multiplier

在两种情况下，被乘数将在一个累加器中，具体取决于被乘数和乘数的大小，并且根据操作数的大小，生成的乘积还将存储在两个寄存器中。以下部分说明了三种不同情况下的MUL指令-

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.

Sr.No.

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.

Arithmetic1

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.

Arithmetic2

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.

Arithmetic3

例

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

例

以下示例将3乘以2，并显示结果-

现场演示

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

编译并执行上述代码后，将产生以下结果-

The result is:
6

DIV / IDIV指令

除法运算生成两个元素-商和余数。如果是乘法，则不会发生溢出，因为使用了双倍长度寄存器来保存乘积。但是，在分割的情况下，可能会发生溢出。如果发生溢出，处理器将产生中断。

DIV(除法)指令用于未签名的数据，而IDIV(整数除法)用于签名的数据。

句法

DIV / IDIV指令的格式-

DIV/IDIV    divisor

股息在累加器中。两条指令都可以使用8位，16位或32位操作数。该操作影响所有六个状态标志。以下部分说明了三种操作数大小不同的除法情况-

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.

Sr.No.

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.

Arithmetic4

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.

Arithmetic5

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.

Arithmetic6

例

以下示例将8除以2。被除数8存储在16位AX寄存器中，除数2存储在8位BL寄存器中。

现场演示

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

编译并执行上述代码后，将产生以下结果-

The result is:
4