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

2. 2 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)

3. 3 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

4. 4 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

5. 5 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

6. 6 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-ZLowercase (61h-Ah) a-z 4X 0 1 0 0 X6X 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,0DFh;converts from lower to upper case  since DFh = 1101 1111b

7. 7 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  OF=1 iff there is a signed overflow

8. 8 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.

9. 9 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, CF=1 (only BL is affected)

10. 10 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, 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

11. 11 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) = BX*32 + BX*4  So add (BX shifted by 5) to (BX shifted by 2)

12. 12 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),CF=0  (the remainder of the division is lost)

13. 13 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

14. 14 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

15. 15 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

16. 16 Rotate with CF  RCL rotates to the left with participation of CF  RCR rotates to the right with participation of CF

17. 17 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

18. 18 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 =

19. 19 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

20. 20 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

21. 21 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 < 10 then convert to ‘0’..’9’ else convert to ‘A’..’F’ } end Repeat  The complete ASM coding is left to the reader

22. 22 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 != ) 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