Question

Assume address of 8-bit x8 is 0000_0000, which contains a5 hex. 1. MOV (0, EAX) 2. MOV (x8, AL)

Now hind value of each of the following: (Answer all in hex, like XX, XXXX, XXXX_XXXX, where x is a hex digit. Include leading zeros) 1. AL ____
2. AH ____
3. AX ____
4. EAX ____

Answer 1

The values of AL, AH, AX, and EAX are determined based on the given memory values and MOV instructions.

Step-by-step explanation:

To determine the values of AL, AH, AX, and EAX given the hexadecimal value a5 in the memory location with address 0000_0000, we need to consider the MOV instructions and their impact on the registers.

1. MOV (0, EAX): This instruction moves 4 bytes from memory address 0 to the EAX register. In hexadecimal, the value at this memory address is a5. Therefore, after the instruction, EAX will hold the value a5 (0000_0000_a5).
2. MOV (x8, AL): This instruction moves 1 byte from the memory location with address x8 to the AL register. In this case, the memory address is 0000_0000. Since the value at this address is a5, after the instruction, AL will hold the value a5.

With this information, we can determine the following:

1. AL = a5
2. AH (higher byte of AX) = 00
3. AX (16-bit register composed of AH and AL) = a500
4. EAX (32-bit register composed of AX) = 0000_0000_a500

Therefore, the values are:

• AL = a5
• AH = 00
• AX = a500
• EAX = 0000_0000_a500

Answer 2

After executing the instruction MOV (x8, AL), the AL register contains A5, AH remains unchanged at 00, AX is 00A5, and EAX ends up with 000000A5, assuming the upper parts of EAX were initially 0.

Step-by-step explanation:

The question involves understanding the x86 architecture and how data is moved into the various parts of the processor's registers. Assuming the memory at address 0x0000 contains the hex value A5, we must evaluate the contents of the EAX register after executing the given instructions.

• The instruction MOV (0, EAX) does not appear to be valid in the context of most assembly languages. It is likely intended to move the value at memory address 0 into EAX, or it may be a typo or misunderstanding. Without clarification, we will focus on the second instruction.
• The instruction MOV (x8, AL) suggests moving the value from memory address labeled 'x8' (or simply the content at memory address 0 since the starting address is 0x0000) into the AL register. Since the value at this address is A5, after this operation, AL would contain A5.

Therefore:

1. AL will contain A5.
2. AH previously contents are not changed by this operation, so if it was initially 00, it would still be 00.
3. AX, which consists of AH and AL, will thus be 00A5.
4. EAX also includes AX, and if the higher order bytes (the E part) were initially 0, then EAX would be 000000A5.