Question

Address ∣ Value

0XA02A ∣ 0X40
0XA02B ∣ 0X00
0XA02C ∣ 0X02
0XA02D ∣ 0XB0

a. (5 points) Assuming that these four bytes represent. an sign-marnitude 32-bit integer in a little-endian machine, its decimal numerical value is ___
b. (5 points) Assuming that these four bytes represent a 2 's complement 32 -bit integer in a big-endian machine, its decimal numerical value is ___

Answer 1

For a little-endian sign-magnitude 32-bit integer, the memory values correspond to -688, and for a big-endian 2's complement 32-bit integer, the values correspond to 1073779696.

Step-by-step explanation:

To answer the student's question, we must interpret the values stored in memory differently depending on the endianness and integer representation. For a little-endian machine, the bytes are read from the least significant to the most significant, while in a big-endian machine, the bytes are read from the most significant to the least significant.

a. Little-Endian Sign-Magnitude: Sign-magnitude representation indicates the first bit is for the sign, and the rest are the magnitude. The bytes in little-endian order are 0XB0, 0X02, 0X00, 0X40. Since the highest bit in 0XB0 is set, this represents a negative number. The magnitude is then 0X00 0X02 0XB0 (ignoring the sign bit), which is hexadecimal for 688. Therefore, the number is -688.

b. Big-Endian 2's Complement: In 2's complement, the value is negative if the highest bit is set. Reading the bytes in big-endian gives us 0X40 0X00 0X02 0XB0. The highest bit is not set, indicating this is a positive number. We simply convert 0X400002B0 to decimal, yielding 1073779696.