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In the following instruction sequence, show the changed values of AL where indic

ID: 3827208 • Letter: I

Question

In the following instruction sequence, show the changed values of AL where indicated, in binary: mov al, 11001011b and al,00111011b;a mov al, 4Dh and al, 69h; b mov al,00101100b or al, 84h; c. mov al, 91h xor al, 57h; d Show the changed values of AL where indicated, in hexadecimal: mov a1, oACh not al; a mov al, 5Bh and al, 44h; b mov al, 96h or al, 52h; c. mov al, 62h xor al, 0B5h;d. Show the values of the Carry, Zero, and Sign flags where indicated: mov al,00110011b test al, 4; a. CF = ZF = SF = mov al, 4 cmp al, 5 b. CF = ZF = SF = mov al, 6 cmp al, 7; c. CF = ZF = SF =

Explanation / Answer

1.a

mov al, 11001011b

AL = 11001011b

and al, 00111011b

AL = (11001011b) AND (00111011b)

    = 00001011b

b.

mov al, 4Dh

AL = 4DH = 01001101B

and al,69H

AL = (01001101B) AND (01101001B)

     = (01001001B)

C.

mov al, 00101100b

AL = 00101100b

or al, 84H

AL = (84H) OR (00101100b)

     = (10000100B) OR (00101100b)

     = (10101100B)

d.

mov al, 91H

AL = 91H = 10010001B

xor al, 57H

AL = (10010001B) XOR (57H)

     = (10010001B) XOR (01010111B)

     = 11000110B

2.a

mov al, 0ACh

AL = 0ACH = 10101100b

not al

AL = NOT(10101100b)

     = 01010011B = 53H

b.

mov al, 5Bh

AL = 5BH = 01011011B

and al, 44h

AL = (01011011b) AND (44H)

     = (01011011B) AND (01000100b)

     = (01000000b) = 40H

C.

mov al, 96h

AL = 96H = 10010110b

or al, 52h

AL = (96H) OR (52H)

     = (10010110b) OR (01010010b)

      = (11010110b) = 0D6H

d.

mov al, 62h

AL = 62H = (0110 0010b)

xor al, 0B5h

AL = 62H XOR 0B5H

      = (0110 0010b) XOR ( 1011 0101b)

     = (1101 0111b) = 0D7H

3.a

mov al, 00110011b

AL = 00110011b

test al, 4

ANDing AL data with 4, result will not store anywhere, but flags will change.

CF = 0, ZF = 1, SF = 0, PF = 1

b.

mov al, 4

AL = 0000 0100B

cmp al, 5

Compare AL data with 5,

result is not stored anywhere, flags will change according to result.

CF = 0, ZF = 0, SF = 1, PF = 0

c.

mov al, 6

AL = 0000 0110b

cmp al, 7

Compare AL data with 7,

result is not stored anywhere, flags will change according to result.

CF = 0, ZF = 0, SF = 1, PF = 0

1.a

mov al, 11001011b

AL = 11001011b

and al, 00111011b

AL = (11001011b) AND (00111011b)

    = 00001011b

b.

mov al, 4Dh

AL = 4DH = 01001101B

and al,69H

AL = (01001101B) AND (01101001B)

     = (01001001B)

C.

mov al, 00101100b

AL = 00101100b

or al, 84H

AL = (84H) OR (00101100b)

     = (10000100B) OR (00101100b)

     = (10101100B)

d.

mov al, 91H

AL = 91H = 10010001B

xor al, 57H

AL = (10010001B) XOR (57H)

     = (10010001B) XOR (01010111B)

     = 11000110B

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