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Hello everyone I have a question for assembly language course and I need a good

ID: 3578751 • Letter: H

Question

Hello everyone I have a question for assembly language course and I need a good expert who can helps me to solve that question. Thank you 3. (5 points) Instruction Sequence. a) Given the following instruction sequence, show the resulting value of the AL register, where indicated, in hex: mov al 0C2h i shr al 1 mov al 0C2h. ii sar al 1 mov al 0C2h iii. sar al, 4 mov al 0C2h iv rol al 1 b) Given the following instruction sequence, show the resulting value of the ALregister, where indicated, in hex: mov al 0C2h i ror al, 3 mov al 0C2h ii rol al, 7 stc mov al, 0C2h iii rcl al, 1 stc mov al, 0C2h iv rcr al 3

Explanation / Answer

AL = 0C2h

0C2h in Binary is 1100 0010

a)
   i. shr => Shift right by 1 bit. shifted bit is lost.
           After shifting 0C2h by 1 bit, we get
                   01100001 = 61
                  
   ii. sar => Shift Arithmetic right by 1 bit.
           Here the sign of the number is repeated on MSB side.
           After shifting 0C2h right by 1 bit, we get
                   11100001 = E1
                  
   iii. sar => Shift Arithmetic right by 4 bits.
               After shifting, we get
                   11111100 = FC
                  
   iv. rol => Rotate left by 1 bit. Here after rotation the shifted bits
               are moved to LSB side.
               After rotation, we get
                   10000101 = 85
                  
          
b)
   i. ror => Rotate right by 3 bits. After rotation, the bits are moved to MSB side.
               After rotation, we get  
                   01011000 = 58
                  
   ii. rol => Rotate left by 7 bits.
               After rotation, we get
                   01100001 = 61
                  
   iii. stc => Set carry bit to 1.
       rcl => Rotate left by 1 bit through carry.
               Moved out bit is transferred to carry and bit at carry is moved to LSB position.
               After rotation, we get
                   Carry = 1
                   10000101 = 85
                  
   iv. rcr => Rotate through carry bit RIGHT by 3 bits.
               Again, carry bit is set through STC instruction.
              
               After rotation, we get  
                   Carry = 1
                   01110000 = 70

                  

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