Given that the IEEE 754 single precision binary representation of 5.375 is 0100 0000 1010 1100 0000 0000 0000 0000, consider this C program on a little endian machine: main() { float a = 5.375; char *p; int i; p = (char *) &a; for (i = 0; i <= 3; i++) printf(" %02x", (unsigned char) p[i]); } What hexadecimal byte sequence will this program print?

Introduction / Context:
This question combines knowledge of IEEE 754 floating point representation with the concept of endianness in memory layout. The program treats a float as an array of bytes and prints those bytes in hexadecimal. To answer correctly, you must understand the bit pattern of 5.375 and how little endian systems store multi byte values in memory.

Given Data / Assumptions:

• The IEEE 754 single precision representation of 5.375 is given as 0100 0000 1010 1100 0000 0000 0000 0000.
• The machine is little endian, meaning that the least significant byte is stored at the lowest memory address.
• The pointer p points to the first byte of the float a, and the loop prints four bytes p[0] to p[3].

Concept / Approach:
The 32 bit pattern 0100 0000 1010 1100 0000 0000 0000 0000 can be grouped into hexadecimal as 0x40AC0000. In big endian order, the bytes in memory would appear as 40 AC 00 00 from lowest to highest address. In little endian order, the byte order is reversed: the least significant byte 00 is stored at the lowest address, followed by 00, then AC, then 40 at the highest of the four addresses. Because p points to the lowest address of a, p[0] is 0x00, p[1] is 0x00, p[2] is 0xAC, and p[3] is 0x40.

Step-by-Step Solution:
Step 1: Convert the given binary pattern to hexadecimal. The bits correspond to 0x40AC0000.Step 2: Identify the four bytes of this value in big endian order: 0x40, 0xAC, 0x00, and 0x00.Step 3: Reverse the byte order for a little endian machine. The sequence in memory from lowest to highest address becomes 0x00, 0x00, 0xAC, 0x40.Step 4: The pointer p points to the first byte, so p[0] is 0x00, p[1] is 0x00, p[2] is 0xAC, and p[3] is 0x40.Step 5: The loop prints each byte with format " %02x", so the output is " 00 00 ac 40" with spaces before each byte. Ignoring the leading space, the sequence is 00 00 ac 40.

Verification / Alternative check:
You can write a small C program on a known little endian platform such as x86, set a float to 5.375, and print its bytes in hex. The sequence 00 00 ac 40 appears on most standard systems. Tools that examine memory layouts and debuggers that display the bytes of a float also confirm this ordering.

Why Other Options Are Wrong:
Option B, 40 ac 00 00, corresponds to big endian order and would be correct only if the machine stored the most significant byte first.Option C and option D represent other permutations of the bytes that do not match the defined little endian layout for this value.

Common Pitfalls:
A frequent mistake is to ignore endianness and assume that the in memory order always matches the hex representation of the number. Another pitfall is to mis group bits when converting from binary to hex. Remember that endianness affects byte ordering in memory, not the bit order inside individual bytes.

Final Answer:
The correct answer is 00 00 ac 40.