5.43 · exercise
Which end is up?
The starter reads one 32-bit word, typed in hex (for example 12345678), into a local in the stack frame. Print the word's four bytes in the order they sit in memory, lowest address first, each as two hex digits.
AArch64 stores a word little-endian: the least significant byte goes at the lowest address. So 12345678 prints as 78 56 34 12. Load one byte at a time from fp + word_s, fp + word_s + 1, and so on; %02x prints a byte as two hex digits, with a leading zero when it needs one.
What is checked
- the four bytes on one line, lowest address first
- the program exits cleanly
- each byte is loaded from memory on its own (
ldrbpresent) - your program also runs on words you do not see, so the bytes have to come from memory
specification
stdin12345678
stdoutprints the right output
exitexits with the right code
sourceuses
ldrbhiddenright output and exit code on 6 more inputs you do not see
We run your program on the input above and on the hidden ones, and compare what it does. Nothing is matched against a stored solution.