Question

For each part of this problem, assume the "before" values when the given instruction is executed. Give the requested "after" values. (The operand are word size to reduce the number of pairs of bits to combine.)

Answer 1

a)

   or instruction will make bitwise OR operation between the 2 operands

   given ax is FA 75 and next operand is fff0

   ax in binary is 1111 1010 0111 0101

   fff0 in binary is 1111 1111 1111 0000

   Hence or operation of the both operands is 1111 1010 0111 0000

   which is FA70 in hexa decimal

   Hence After execution of the instruction

   AX : FA70h

   SF : 1 (as the MSB of the result is 1 it indicates negative number)

   ZF : 0 (as the result is non zero)

b)

   TEST instruction performs a bitwise AND on two operands

   it sets the flags CF and OF to zero, SF is set to the most significant bit of the result

   If the result is 0, the ZF is set to 1

   given ax is FA 75 and next operand is 0004

      ax in binary is 1111 1010 0111 0101

   0004 in binary is 0000 0000 0000 0100

   Hence and operation of the both operands is 0000 0000 0000 0100

   which is 0004 in hexa decimal

   Hence After execution of the instruction

   AX : 0004h

   SF : 0 (as the MSB of the result is 0 it indicates positive number)

   ZF : 0 (as the result is non zero)

c)

   xor instruction performs a bitwise XOR on two operands

   given dx is B6 A3 and next operand is 64C8

      dx in binary is 1011 0110 1010 0011

   64C8 in binary is 0110 0100 1100 1000

   Hence xor operation of the both operands is 1101 0010 0110 1011

   which is D26B in hexa decimal

   Hence After execution of the instruction

   AX : D26Bh

   SF : 1 (as the MSB of the result is 1 it indicates negative number)

   ZF : 0 (as the result is non zero)

d)

   sal instruction performs shift arithimetic left on two operands

   it will shift left by the number of bit positions equal to count and fill the vacted bit positions

   on the right with 0's

   given ax is A8 B5 and cl is 04 which means we have to shift A8 B5 by 4 positions left

   ax in binary is 1010 1000 1011 0101

   after shifting by 4 positions to left

   ax will be 1000 1011 0101 0000

   which is 8B50 in hexa decimal

   Hence After execution of the instruction

   AX : 8B50

   CF : 0 (as the 4th bit shifted to the left in AX is 0 (in the nibble 1010))

e)

   rcr instruction performs rotate right through carry on two operands

   given ax is A8 B5 and count is 01 which means we have to rotate A8 B5 by 1 position right through carry.

   ax in binary is 1010 1000 1011 0101

   after rotating A8 B5 by 1 position right through carry.

   ax will be 0101 0100 0101 1010

   which is 545A in hexa decimal

   Hence After execution of the instruction

   AX : 545A

   CF : 1 (as the 1st bit rotated to the right in AX is 1)

f)

   sar instruction performs shift arithimetic right on two operands

   it will shift right by the number of bit positions equal to count and fill the vacted bit positions

   on the left with 0's

   given BX is 8E BA and cl is 03 which means we have to shift 8E BA by 3 positions right

   BX in binary is 1000 1110 1011 1010

   after shifting by 3 positions to right

bx will be 0001 0001 1101 0111

   which is 11D7 in hexa decimal

   Hence After execution of the instruction

   BX : 11D7

   CF : 0 (as the 3rdbit shifted to the right in BX is 0 )