2 ms·
> (assuming 'x' is kept live), and the "load" of 'y' into re-computing y from x If I understand correctly, what you describe is called rematerialization (https
by tom_mellior 9y ago
> (assuming 'x' is kept live), and the "load" of 'y' into re-computing y from x
If I understand correctly, what you describe is called rematerialization (https://en.wikipedia.org/wiki/Rematerialization https://en.wikipedia.org/wiki/Rematerialization), and yes, it's standard. As with everything else in register allocation, it's difficult to decide when and how to do it.
- kannanvijayan 9y agoCool! Thanks for the info and correct terminology.