Why are only 8 bytes of the hash compared with the target?

Question

This is the line in xmr-stak where the hash is compared to the target:

if (*piHashVal < oWork.iTarget)
executor::inst()->push_event(ex_event(result, oWork.iPoolId));

However, if you look above, *piHashVal is defined as:

piHashVal = (uint64_t*)(result.bResult + 24);

Where result.bResult is the whole hash (a 32 byte long array), and it's just using the last 8 bytes of it.

Why not compare all 32 bytes to the target? And why is it comparing only the last 8 bytes? Does this mean that the code is little endian only?

Answer 1

It is little endian, and it compares only 8 bytes because it's lazy and if this matches, then the other 24 bytes don't matter (they'd matter if those 8 bytes were equal, in which case you're deemed not to have found a share, where you could have found one). This is probably done for some combination of speed and laziness.