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1 Logic, Shift, and Rotate Logic, Shift, and Rotate Instructions Instructions Read Sections 6.2, 7.2 and 7.3 of textbook Read Sections 6.2, 7.2 and 7.3 of textbook

# 1 Logic, Shift, and Rotate Instructions Read Sections 6.2, 7.2 and 7.3 of textbook

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Logic, Shift, and Rotate InstructionsLogic, Shift, and Rotate Instructions

Read Sections 6.2, 7.2 and 7.3 of textbookRead Sections 6.2, 7.2 and 7.3 of textbook

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Logic InstructionsLogic 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.)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.)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 ECX=0 we can do:OR ecx,ecx

since this does not change the number in ECX and set ZF=1 if and only if ECX=0

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Logic Instructions (cont.)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.)*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 X5X 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,0DFh;converts from lower to upper case

since DFh = 1101 1111b

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Logic Instructions (cont.)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 useNEG destination

affect SF and ZF according to result CF is set to 1 unless the result is 0 OF=1 iff there is a signed overflow

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Exercise 1Exercise 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 LeftShifting 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, CF=1 (only BL is affected)

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Shifting Multiple Times to the LeftShifting 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, CF=0 Effect on OF for all shift and rotate instructions (left and

right): For any single-bit shift/rotate: OF=1 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 MultiplicationFast 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) = BX*32 + BX*4 So add (BX shifted by 5) to (BX shifted by 2)

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Shifting bits to the rightShifting 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),CF=0

(the remainder of the division is lost)

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Arithmetic Shift SARArithmetic 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)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 ROLExamples 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 CFRotate 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: 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 2Exercise 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 OutputApplication: 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 InputApplication: 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 entrynext:getchcmp eax, 10 ;end of input line reached?je exit ;yes then exitand al, 0Fh ;no, convert to binary valueshl ebx, 1 ;make room for new valueor bl, al ;put value in ls bitjmp 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 OutputAlgorithm for Hex Output

To display in hexadecimal the content of EAXRepeat 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 < 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 InputAlgorithm 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 != <CR>) 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