I got asked this question in my Computer Organization and Architecture assesment. Can you guys please help me with this?
"A processor is designed to receive a data stream in little endian format and store it at address starting DATA in word order. If the received data is AABBCCDDEE...., show the layout of storage in memory."
I didn't get proper answer for this question, i thought the size of the bit of the computer matters(whether it is 32 or 64 bits). If yes, assume it is 32 bit computer. I am not sure of the answer. I think the answer for this is to write the character in reverse can someone please explain how and why?
On a little-endian system, the least significant byte goes to the lowest memory address. If you receive a byte stream in little-endian order (e.g., AA BB CC DD EE ...) and store it word by word in a 32-bit processor, you group the bytes in chunks of four. For each 4-byte group (word), the first byte is the least significant byte, so it ends up at the smallest address within that word.
Concretely, for the first word AA BB CC DD:
AA goes to DATA + 0BB goes to DATA + 1CC goes to DATA + 2DD goes to DATA + 3So, in the actual byte-by-byte memory dump (from lowest to highest address), you’ll see AA BB CC DD EE FF GG HH ... exactly in the order the bytes arrive. But if you interpret AA BB CC DD as a 32-bit value in little-endian format, it corresponds to the integer 0xDDCCBBAA in typical hex notation (because the “DD” byte is the most significant part of that integer, even though it’s in the highest memory address of the word). That “reverse” effect often confuses people, but it’s really just how little-endian works at the per-word level.