Thread: Splitting up int16_t to two int8_t

Originally Posted by Exile

Does taht same problem exist for the solutions that use bit shifts? The shifts would work, but the results of the shift are switched.

That was my point: Bit shifts (or arithmetic with '/' and '%' operators) allow you to extract the bytes whatever order you require. (Not endian-dependent.)

The shift and arithmetic operators work on numbers in registers. Upper bits are upper bits; lower bits are lower bits.

When the individual bytes get stored in memory, the byte order is dependent on the endianness of the machine. This shows up with different results for bid-endian systems and little-endian systems using the union.

Regards,

Dave