Question 54

Let the 2-byte integer stored in memory at addresses $A$ and $A+1$ consist of bytes $B_0$ and $B_1$, respectively.

In a Little Endian (LE) architecture, the byte at the lower address is the least significant byte (LSB), and the byte at the higher address is the most significant byte (MSB):

In a Big Endian (BE) architecture, the byte at the lower address is the most significant byte (MSB), and the byte at the higher address is the least significant byte (LSB):

We are given that $V_{\text{LE}} - V_{\text{BE}} = 255$:

Thus, for an unsigned integer on a little-endian computer, its high-order byte $B_1$ must exceed its low-order byte $B_0$ by exactly 1.

Check each option represented in hexadecimal as 0x[MSB][LSB], where MSB is $B_1$ and LSB is $B_0$:

• Option A: 0x6665: $B_1 = 0x66 = 102$, $B_0 = 0x65 = 101$. Here $B_1 - B_0 = 102 - 101 = 1$. This satisfies the condition!
• Option B: 0x0001: $B_1 = 0x00$, $B_0 = 0x01$. Here $B_1 - B_0 = -1 \ne 1$.
• Option C: 0x4243: $B_1 = 0x42$, $B_0 = 0x43$. Here $B_1 - B_0 = -1 \ne 1$.
• Option D: 0x0100: $B_1 = 0x01$, $B_0 = 0x00$. Here $B_1 - B_0 = 1 - 0 = 1$. This satisfies the condition!

Therefore, choices A and D are correct.