문제는 : how how the value 0xabcdef12 would be arranged in memory of a little-endian and a bigendian machine. Assume the data are stored starting a address 0 and that the word size is 4 bytes.


그래서 답을 


big-endian
(memory)
low ---------------------> high
ab | cd | ef | 12

little-endian
(memory)
low ---------------------> high
12 | ef | cd | ab

이렇게 적었는데. 아니라네여...ㅠㅠ\

답이 뭐죠?ㅠㅠ