Logical operation instruction

Link to the original text —— SevenOne’Blog

Cyclic shift

1. Shift is to shift the binary number in the register .
2. stay debug The values in the registers in the mode have 16 Base display .
3. When only one position is moved , You can put numbers 1 Put it on the source operand of the move instruction . If the number of moving digits is greater than 1, You need to save the moving digits to cl In the register .

Move left ROL

1. Move the highest weight to the lowest full-time weight .

2. example ：0110 The result of moving left is 1100

3. Code implementation

   • demand ： Yes 12h Move one bit to the left
   • Running results ：24h.
   • principle ：12h = 00010010b, After shifting to the left 00100100 namely 24h

   code segment
       assume cs:code
       main:
           mov bx,12h
           rol bx,1
           
       edit:
           mov ah,4ch
           int 21h
   code ends
       end main
   

   

Move right ROR

1. Move the lowest weight to the highest full-time weight .
2. example ：0110 The result of moving right is 0011

Logical operations

Knowledge point

1. Logical operation with high-level language , Follow the and or principle .
2. And operation ： The whole is true , A fake is a fake .
3. Or operations ： One is true , All false is false .
4. Bitwise operation refers to the operation in binary bits .
5. In binary ,1 Said really ,0 Said the false .

Instructions

1. Bitwise and operation ：AND
2. Bitwise OR operation ：OR
3. Bitwise non operation ：NOT
4. Bitwise exclusive or operation ：XOR

Example

1. Calculation 24h and 36h And operations

   code segment
       assume cs:code
       main:
           mov bx,24h
           and bx,36h
   
       edit:
           mov ah,4ch
           int 21h
   code ends
       end main
   

   

2. Calculation 24h and 36h Or operation of

   code segment
       assume cs:code
       main:
           mov bx,24h
           or bx,36h
   
       edit:
           mov ah,4ch
           int 21h
   code ends
       end main
   

   

3. Keep the end number of binary , Other high bits are cleared .

   code segment
       assume cs:code
       main:
           mov bx,0ffffh
           and bx,01h
   
       edit:
           mov ah,4ch
           int 21h
   code ends
       end main
   

   

Comprehensive application

1. Reserved bits and 1 Do with the operation .

2. Zero position sum 0 Do with the operation .

Output binary

1. demand ： Output the value in the register to the terminal in binary form .

2. Code

   code segment
       assume cs:code
       main:
           mov bl,12h
           mov cx,8h
           lopi:
               rol bl,1
               mov dl,bl
               and dl,01h
               add dl,30h
               call d02
           loop lopi
   
           mov dl,42h
           call d02
   
       edit:
           mov ah,4ch
           int 21h
           
       d02 proc near
           mov ah,02h
           int 21h
           ret
       d02 endp
   code ends
       end main 
   

   

Output hex

1. demand ： Output the value in the register to the terminal in hexadecimal form .

2. Code

   code segment
       assume cs:code
       main:
           mov bl,0a2h
           mov cx,2
           lopi:
               push cx
               mov cl,4
               rol bl,cl
               pop cx
   
               mov dl,bl
               and dl,0fh
               cmp dl,09h
               jg letter
               jng number
   
               letter:
                   add dl,37h
                   call d02
                   jmp endcmp
   
               number:
                   add dl,30h
                   call d02
   
               endcmp:
           loop lopi
   
           mov dl,48h
           call d02
   
       edit:
           mov ah,4ch
           int 21h
       d02 proc near
           mov ah,02h
           int 21h
           ret
       d02 endp
   code ends
       end main