Logic, Shift, and Rotate Instructions

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Logic, Shift, and Rotate Instructions

• Chapter 4

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Logic Instructions

• Syntax for AND, OR, XOR, and TEST instructions
• op-code destination, source
• They perform the Boolean bitwise operation and
  store the result into destination.
• TEST is just an AND but the result is not stored.
  TEST affects the flags just like AND does.
• Both operands must be of the same type
• either byte, word or dword
• Both operands cannot be mem
• again mem to mem operations are forbidden
• They clear (ie put to zero) CF and OF
• They affect SF and ZF according to the result of
  the operation (as usual)

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Logic Instructions (cont.)

• The source is often an imm operand called a bit
  mask used to fix certain bits to 0 or 1
• To clear a bit we use an AND since
• 0 AND b 0 (b is cleared)
• 1 AND b b (b is conserved)
• Ex to clear the sign bit of AL without affecting
  the others, we do
• AND al,7Fh msb of AL is cleared
• since 7Fh 0111 1111b

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Logic Instructions (cont.)

• To set (i.e fix to 1) certain bits, we use OR
• 1 OR b 1 (b is set)
• 0 OR b b (b is conserved)
• To set the sign bit of AH, we do
• OR ah,80h
• To test if ECX0 we can do
• OR ecx,ecx
• since this does not change the number in ECX and
  set ZF1 if and only if ECX0

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Logic Instructions (cont.)

• XOR can be used to inverse certain bits
• b XOR 1 NOT(b) (b is complemented)
• b XOR 0 b (b is conserved)
• Ex to initialize a register to 0 we can use
• XOR ax,ax
• Since b XOR b 0 (b is cleared)
• This instruction uses only 2 bytes of space.
• The next instruction uses 3 bytes of space
• MOV ax,0
• Compilers prefer the XOR method

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Logic Instructions (cont.)

• To convert from upper case letter to lower case
  we can use the usual method
• ADD dl,20h
• But 20h 0010 0000b and bit 5 is always 0 for
  chars from A (41h) to Z (5Ah).
• Uppercase (41h-5Ah) A-Z Lowercase (61h-Ah) a-z
• 4X 0 1 0 0 X 6X 0 1 1 0 X
• 5X 0 1 0 1 X 7X 0 1 1 1 X
• Hence, adding 20h gives the same result as
  setting this bit 5 to 1. Thus
• OR dl,20h converts from upper to lower case
• AND dl,0DFhconverts from lower to upper case
• since DFh 1101 1111b

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Logic Instructions (cont.)

• To invert all the bits (ones complement), we use
• NOT destination
• does not affect any flag and destination cannot
  be an imm operand
• Recall that to perform twos complement, we use
• NEG destination
• affect SF and ZF according to result
• CF is set to 1 unless the result is 0
• OF1 iff there is a signed overflow

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Exercise 1

• Use only one instruction among AND, OR, XOR, and
  TEST to do the following task
• (A) Convert the ASCII code of a decimal digit
  ('0 to '9) contained in AL to its numerical
  value.
• (B) Fix to 1 the odd numbered bits in EAX (ie
  the bits numbered 1, 3, 5) without changing the
  even numbered bits.
• (C) Clear to 0 the most significant bit and the
  least significant bit of BH without changing the
  other bits.
• (D) Inverse the least significant bit of EBX
  without changing the other bits.

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Shifting Bits to the Left

• To shift 1 bit to the left we use
• SHL dest,1
• each bit is shifted one position to the left
• the lsb (least significant bit) is filled with 0
• the msb (most significant bit) is moved into CF
  (so the previous content of CF is lost)
• dest can be either byte, word or dword
• Example
• mov bx,80h BX 0080h
• shl bl,1 BX 0000h, CF1 (only BL is affected)

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Shifting Multiple Times to the Left

• Two forms are permitted
• SHL dest,CL CL number of shifts
• SHL dest, imm8
• SHL affects SF and ZF according to the result
• CF contains the last bit shifted out
• mov bh,82h BH 1000 0010b
• shl bh,2 BH 0000 1000b, CF0
• Effect on OF for all shift and rotate
  instructions (left and right)
• For any single-bit shift/rotate OF1 iff the
  shift or rotate changes the sign bit
• For multiple-bit shift/rotate the effect on OF
  is undefined
• Hence, sign overflows are signaled only for
  single-bit shifts and rotates

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Fast Multiplication

• Each left shift multiplies by 2 the operand for
  both signed and unsigned interpretations. Ex
• mov ax,4 AX 0004h
• mov bx,-1 BX FFFFh
• shl ax,2 AX 0010h 16
• shl bx,3 BX FFF8h -8
• Multiplication by shifting is very fast. Try to
  factor your multiplier into powers of 2
• BX 36 BX (32 4) BX32 BX4
• So add (BX shifted by 5) to (BX shifted by 2)

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Shifting bits to the right

• To shift to the right use either
• SHR dest,CL value of CL number of shifts
• SHR dest,imm8
• the msb of dest is filled with 0
• the lsb of dest is moved into CF
• Each single-bit right shift divides the unsigned
  value by 2. Ex
• mov bh,13 BH 0000 1101b 13
• shr bh,2 BH 0000 0011b 3 (div by 4),CF0
• (the remainder of the division is lost)

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Arithmetic Shift SAR

• Is needed to divide the signed value by 2
• SAR dest,CL value of CL number of shifts
• SAR dest,imm8
• the msb of dest is filled with its previous value
  (so the sign is preserved)
• the lsb of dest is moved into CF
• mov ah,-15 AH 1111 0001b
• sar ah,1 AH 1111 1000b -8
• the result is rounded to the smallest integer
• (-8 instead of -7)
• in contrast
• shr ah,1 gives ah 0111 1000b 78h

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Rotate (without the CF)

• ROL rotates the bits to the left (same syntax)
• CF gets a copy of the msb

• ROR rotates the bits to the right (same syntax)
• CF gets a copy of the lsb

• CF reflect the action of the last rotate

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Examples of ROL

• mov ah,40h ah 0100 0000b
• rol ah,1 ah 1000 0000b, CF 0
• rol ah,1 ah 0000 0001b, CF 1
• rol ah,1 ah 0000 0010b, CF 0
• mov ax,1234h ax 0001 0010 0011 0100b
• rol ax,4 ax 2341h
• rol ax,4 ax 3412h
• rol ax,4 ax 4123h

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Rotate with CF

• RCL rotates to the left with participation of CF

• RCR rotates to the right with participation of CF

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Ex inverting the content of AL

• Ex whenever AL 1 1 0 0 0 0 0 1b we want to
  have
• AL 1 0 0 0 0 0 1 1b
• mov ecx,8 number of bits to rotate
• start
• shl al,1 CF msb of AL
• rcr bl,1 push CF into msb of BL
• loop start repeat for 8 bits
• mov al,bl store result into AL

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Exercise 2

• Give the binary content of AX immediately after
  the execution of the each instruction below
  (Consider that AX 1011 0011 1100 1010b before
  each of these instructions)
• (A) SHL AL,2 AX
• (B) SAR AH,2 AX
• (C) ROR AX,4 AX
• (D) ROL AX,3 AX
• (E) SHL AL,8 AX

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Application Binary Output

• To display the binary number in EAX
• MOV ECX,32 count 32 binary chars
• START
• ROL EAX,1 CF gets msb
• JC ONE if CF 1
• MOV EBX,0
• JMP DISP
• ONE MOV EBX,1
• DISP PUTCH EBX
• LOOP START

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Application Binary Input

• To load EAX with the numerical value of a binary
  string (ex 101100101...) entered at the
  keyboard
• xor ebx,ebx clear ebx to hold entry
• next
• getch
• cmp eax,10 end of input line reached?
• je exit yes then exit
• and al,0Fh no, convert to binary value
• shl ebx,1 make room for new value
• or bl,al put value in ls bit
• jmp next
• exit
• mov eax,ebx eax holds binary value
• In AL we have either 30h or 31h (ASCII code of
  0 and 1)
• Hence, AND AL,0Fh converts AL to either 0h or 1h
• Hence, OR BL,AL possibly changes only the lsb of
  BL

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Algorithm for Hex Output

• To display in hexadecimal the content of EAX
• Repeat 8 times
• ROL EAX,4 the ms 4bits goes into ls 4bits
• MOV DL,AL
• AND DL,0Fh DL contains num value of 4bits
• If DL lt 10 then convert to 0..9
• else convert to A..F
• end Repeat
• The complete ASM coding is left to the reader ?

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Algorithm for Hex Input

• To load EAX with the numerical value of the
  hexadecimal string entered at the keyboard
• XOR EBX,EBX EBX will hold result
• While (input char ! ltCRgt) DO
• convert char into numerical value
• left shift EBX by 4 bits
• insert value into lower 4 bits of EBX
• end while
• MOV EAX,EBX
• The complete ASM coding is left to the reader ?