Microcomputer principle and interface technology exercise 1

Reference books ：《 Microcomputer principle and interface technology （ The third edition ）》 Hongyongqiang
The original title is in the 125 page ,4-12
Only provide personal ideas , Maybe the solution is not very good , Welcome to exchange .

subject

  12. Programming , No multiplication Z=10×X+Y/8, Use shift operation .

analysis

  ① First of all, this is a simple calculation problem , There is addition 、 Multiplication 、 Division operations . Multiplication is not allowed , Use shift instead of multiplication . So we can only use addition 、 division 、 displacement .
  ② Convert multiplication to shift .
     The computer stores binary data , Moving one bit to the left is equivalent to doing X2 Arithmetic . Moving one bit to the right is equivalent to doing ÷2 Arithmetic .
     But not exactly the same , It also depends on the instructions used , Involving overflow , Cyclic shift, etc .
  ③ If 10xX, There are two ways to calculate , One is the X Split into 2 Of n1 Power +2 Of n2 Power +… Another way is to put 10 Split into 2 Of 3 Power +2 Of 1 Power . Solve the existing problems as simple as possible , Here choose the latter method .
  ④Y/8 You can divide , But right shift is used here 3 position （2^3=8） Faster .
  ⑤ Finally, add , Come to the conclusion .

flow chart

Instructions needed

   Add instruction ：
    1) Add instruction without carry ADD(add)
     Format ：ADD DEST, SRC
     function ：DEST←SRC+DEST
    2) Add instruction with carry ADC(add with carry)
     Format ：ADC  DEST, SRC
     function ：DEST←SRC+DEST+CF
    ADC It is mainly used in multi byte addition .

   Shift instructions ：
     Logic shift left ：SHL DEST,COUNT
     Logical shift right ：SHR DEST,COUNT
     among , DEST You can use any addressing method other than immediate ; COUNT Indicates the number of shifts , Its value is or equal to 1, Or greater than 1, Greater than 1 The value should be sent to CL register , And then move it .

Program

;4-12   Shift without multiplication  Z=10*X+Y/8
DATA   SEGMENT; Data segment 
       X DW 90 ; The assumption is 90
       Y DW 80 ; hypothesis 80
       Z DW 0  ; The result should be 910
DATA   ENDS     

STACKS SEGMENT; stack segment  
STACKS ENDS      

CODE   SEGMENT; Code segment   
       ASSUME CS:CODE,DS:DATA,SS:STACKS
START: MOV AX,DATA
       MOV DS,AX      
       MOV AX,STACKS
       MOV SS,AX       ; initialization 

       MOV CL,3        ; The number of shifts is 3
       MOV AX,X
       SHL AX,1        ;X Move left 1 position 
       MOV BX,X
       SHL BX,CL       ;X Move left 3 position 
       ADD AX,BX       ;X*2+X*8
       MOV BX,Y      
       SHR BX,CL       ;Y Move right 3 position 
       ADD AX,BX       ;X*10+Y/8
       MOV Z,AX
CODE ENDS
     END START       
   

debugging

   from emu8086 Call up the variable box in , Start running program .