3 ms·
The format is little-endian, so it's necessary to use a little-endian load instruction, which almost all architectures (including most big-endian architectures)
by kentonv 10y ago
The format is little-endian, so it's necessary to use a little-endian load instruction, which almost all architectures (including most big-endian architectures) have.
(But every widely-used arch these days is LE.)