Debug through each line of code and explain the register content and flags.

(We already answered line 10 to 13 for your reference. Start writing your answer from Line 14)

Line: 10

Instruction: mov eax, 12345678h

Register value: EAX = 12345678

Explanation: 12345678 is a hexadecimal value which is 32-bit in binary. EAX register is also 32-bit.

Line 11:

Instruction: mov ax, 1122h

Register value: EAX = 12341122h

Explanation: 1122 is hexadecimal and it is 16-bit in binary. this mov instruction only updates AX (16 bit) register, a part of EAX register. Thats why you can see that the upper portion of EAX register is NOT updated.

Line 12:

Instruction: mov bl, al

Register value: EBX = _ _ _ _ _ _ 22

Explanation: AL register is 8-bit long. When you mov the content of al register (22) to BL register, it only updates the first 8-bit of the EBX register. The rest contains the garbage value.

Line 13:

Instruction: mov bl, ah

Register value: EBX = _ _ _ _ _ _ 11

Explanation: Ah register is 8-bit long. When you mov the content of AH register (11) to BL register, it only updates the first 8-bit of the EBX register. The rest contains the garbage value.

Line 14:

Instruction: mov al, 89h

What Register value of EAX register, after executing line 14.

Explain the content of the EAX register.

Line 15:

Instruction: add al, 10h

What Register value of EAX, after executing line 15?

Do you see any change in flags?

Show the step of the hexadecimal addition.

Line 16:

Instruction: sub al, al

What Register value of EAX, after executing line 16?

Do you see any change in flags?

Show the step of the hexadecimal subtraction.

Line 17, 18:

Instruction:

mov al, 98h

add al, 89h

What Register value of EAX, after executing line 17 and 18?

Do you see any change in flags?

Show the step of the hexadecimal addition.

these are the questions. help me please

Registers EAX - 12341121 EBX00801011 ECX = 887A1005 EDX - 097A1805 ESI = 807A1005 EDI - 097A1005 EIP - 807A1027 ESP - BOSAF908 EBP 805AF914 EFL - BOBBA17 OV - 1 UP - @ EI = 1 PL = ZR = BAC = 1 PE = 1 CY = 1 100. Diagnostic Tools + Diagnostics session: Events LI Process Memory (K 488 Source am X X Source asm 1 386 2 model flat, stdcall 3 stack 4096 4 5 Exit Process PROTO, dwExit Code: DWORD 6 7 .code main PROC 10 OV eax, 12345678 Ovax, 1122h 12 mov bl, 1 nov bl, ah 14 mov al. 89 15 add al, 10 16 sub al, al 17 moval, 981 add al, 89h 19 20 21 invoke Exit Process, in ENOP END sain CPU (% of all proces 100 0 Summary Events M Y Filter - Filt Event Step: Source.am Step: Source asm Step: Source asm Step: Source asm Step: Source asm Step: Source asm 100% No isso found L 21 Chil TABS CRL O Ready Type here to search o EP Registers EAX - 12341121 EBX00801011 ECX = 887A1005 EDX - 097A1805 ESI = 807A1005 EDI - 097A1005 EIP - 807A1027 ESP - BOSAF908 EBP 805AF914 EFL - BOBBA17 OV - 1 UP - @ EI = 1 PL = ZR = BAC = 1 PE = 1 CY = 1 100. Diagnostic Tools + Diagnostics session: Events LI Process Memory (K 488 Source am X X Source asm 1 386 2 model flat, stdcall 3 stack 4096 4 5 Exit Process PROTO, dwExit Code: DWORD 6 7 .code main PROC 10 OV eax, 12345678 Ovax, 1122h 12 mov bl, 1 nov bl, ah 14 mov al. 89 15 add al, 10 16 sub al, al 17 moval, 981 add al, 89h 19 20 21 invoke Exit Process, in ENOP END sain CPU (% of all proces 100 0 Summary Events M Y Filter - Filt Event Step: Source.am Step: Source asm Step: Source asm Step: Source asm Step: Source asm Step: Source asm 100% No isso found L 21 Chil TABS CRL O Ready Type here to search o EP