4 ms·
If it is not on the hot path, it is likely free, but not guaranteed. If it is on the hot path then it is wasting a whole cycle. And of course in highly ALU-depe
by NohatCoder 4y ago
If it is not on the hot path, it is likely free, but not guaranteed. If it is on the hot path then it is wasting a whole cycle. And of course in highly ALU-dependent code it is another instruction, so a fraction of a clock.
- kaba0 4y agoWhat do you mean it wastes a whole cycle? It may indeed have worse performance due to blowing the instruction cache, but I don’t see why would out-of-order execution be slower on the hot path - I doubt there would be too many hot paths without any dependence on memory fetches outside specific benchmarks - the memory loads will take significantly more time even if they hit cache.