CODEWITHSUNDEEP
assemblyrotationshift
hajile
hajile

Reputation: 37

Assembly Shift/Rotate

These are the problems:

I have this hw, but I only know basic knowledge. It will be a great help if anyone can give me a hint on how to finish these problems. I know this may be a desperate move, but atleast I will learn how to answer these kind of problems.

Upvotes: 1

Views: 794

Answers (4)

Sep Roland
Sep Roland

Reputation: 39166

  1. To clear the most significand nibble and set the least significand nibble of AX you can write:

    and ax, 0FFFh  ;Clears the top nibble
or  ax, 000Fh  ;Sets the bottom nibble

    Without and/or, using shifts:

    shl ax, 4      ;Moves the middle part into AH
mov al, 0F0h   ;Prepares for a set low nibble
shr ax, 4      ;Puts everything in place

  2. To calculate 10*AX using shifts:

    shl ax, 1      ;Calculate 2*AX
mov dx, ax     ;Store in DX
shl ax, 2      ;Continue calculating 8*AX
add ax, dx     ;Makes AX = 8*AX + 2*AX

Upvotes: 0

Tommylee2k
Tommylee2k

Reputation: 2731

"and" returns 1 only, if both input bits are 1.
0 and 0 is 0
1 and 0 is 0
0 and 1 is 0
1 and 1 is 1

so "clearing" a bit can be done by "anding" with 0. whatever the bit (b) is, b and 0 is 0

"or" returns 1, if either of the input bits is 1.
0 or 0 is 0
1 or 0 is 1
0 or 1 is 1
1 or 1 is 1

"setting" a bit can be done by "ORing" with 1. whatever the bit (b) is, b or 1 is 1

AX is 16 bit if you AND AX with 0000.1111.1111.1111b (= 0x0FFF ) the topmost nibble (4 bits) will be cleared
(AND with 1 will not change a bit, it remains 1 if it's set, and remains unset if it's 0, so all nibbles that should not be changed are ANDed with 1)

if you OR AX with 0000.0000.0000.1111b ( = 0x000F ) the lower nibble will all be set.
(OR with 0 will not change a bit, it remains 1 if it's set, and remains unset if it's 0, so all other nibbles are ORed with 0)

Upvotes: 1

X0r
X0r

Reputation: 111

The best way to understand these operations is to walk through them step by step. I'll try to explain the process as best I can. With each step I'll include the status of both registers so you can follow along. In order to maintain the upper values of EAX, we have to first move its contents into another register for manipulation.

Start:
EAX: 0011 1100 1001 1111 1011 1100 1001 1111

mov ebx, eax ; copies contents of eax into ebx

EAX: 0011 1100 1001 1111 1011 1100 1001 1111
EBX: 0011 1100 1001 1111 1011 1100 1001 1111

shl ebx, 24 ; this clears the bits off ebx that we're not concerned with
shr ebx, 24 ; shifts back to initial position

EAX: 0011 1100 1001 1111 1011 1100 1001 1111
EBX: 0000 0000 0000 0000 0000 0000 1100 1001

shl ebx, 4 ; shift to the left by 4 bits. bx is where we want it for the next step.

EAX: 0011 1100 1001 1111 1011 1100 1001 1111
EBX: 0000 0000 0000 0000 0000 1100 1001 0000

add bl, 0Fh ; adds 0xF to bl. Or, in binary, 1111. least significant nibble set.

EAX: 0011 1100 1001 1111 1011 1100 1001 1111
EBX: 0000 0000 0000 0000 0000 1100 1001 1111

xor ax, ax ; clears contents of ax, making room for what's in bx

EAX: 0011 1100 1001 1111 0000 0000 0000 0000
EBX: 0000 0000 0000 0000 0000 1100 1001 1111

add ax, bx ; combine the results of bx back into eax for the end result

EAX : 0011 1100 1001 1111 0000 1100 1001 1111

And there you have it! Now that you're familiar with shift operations, I'll leave the last part to you.

Cheers!

Upvotes: 0

Tommy
Tommy

Reputation: 100602

You can set and clear parts of an integral value using bit logic — logical AND and OR in that case.

Superficial examples: I have the bit value 0101; I want to set the lowest two bits. So I OR with 0011. Now I want to clear the two lowest bits. So I AND with 1100.

Upvotes: 0

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