2 ms·
> A classic compiler optimisation is to fuse these two loops together, so that you only iterate over a once Honest question: why is that true? If x is the cost
by ngruhn 6mo ago
> A classic compiler optimisation is to fuse these two loops together, so that you only iterate over a once
Honest question: why is that true? If x is the cost of running one iteration of the first loop and y is the cost for the second loop then the total cost is:
n*x + n*y
After fusing, the cost is:
n*(x+y)
which is the same.
- rowanG077 6mo agoBecause two loops is really: n*(x + c) + n*(y + c) => n*x + n*y + 2*n*c Where c is some loop overhead, such as loading the arguments and work variables into registers and storing them back, checking the loop end condition, increasing the counters etc. So if you fuse you get: n*x + n*y + n*c