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Recall the two methods given in this section for computing (value mod 32 ) when value is an unsoz integer in the EAX register: mov edx,0, intend value to quadword mov ebx,32; divisor div ebx ; divide value by 32 and mov edx, eax ; copy value to DX and edx,0000001fh; compute value mod 32 Find the total number of bytes of object code necessary for each of these methods.

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Answer:

In general, the first method is more versatile because it can handle division by any arbitrary divisor, not just 32. However, it is also more computationally expensive because it involves both a division and a multiplication operation, which can be slow. On the other hand, the second method is more specialized to the case of computing a modulus by a power of 2 (in this case, 32), and as a result, it is more efficient in terms of the number of instructions and the execution time.

Step-by-step explanation:

Let's assume that the instructions are written in x86 assembly language.

Method 1:

The first method involves using the DIV instruction to divide the value by 32, and then copying the quotient to the EDX register, which gives us the result of the integer division. Finally, we use the AND instruction to compute the remainder, which is the same as value mod 32.

The object code for this method would be:

mov edx, 0 ; initialize EDX to 0

mov eax, value ; move value into EAX

mov ebx, 32 ; divisor

div ebx ; divide EAX by EBX, quotient in EAX, remainder in EDX

mov eax, edx ; copy remainder to EAX

and eax, 1Fh ; compute value mod 32

The total number of bytes of object code for this method would be 15 bytes.

Method 2:

The second method involves using the AND instruction directly to compute the remainder, which is the same as value mod 32.

mov eax, value ; move value into EAX

and eax, 1Fh ; compute value mod 32

mov edx, 0 ; clear EDX (optional, depending on context)

The total number of bytes of object code for this method would be 7 bytes.

Therefore, the second method is more efficient in terms of object code size, requiring only 7 bytes compared to the 15 bytes required by the first method.

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