Inline expansion and temporal locality
in [General Programming]
|
Prev: Brand Watches Ebel Women's Beluga Manchette Watch #9057A21-5635406 Discount, Replica, Fake
Next: Tissot Men's PRC 100 (Quartz Gent) Stainless Steel Watch - Replica Watch Fake
From: Stephan Ceram on 19 Apr 2008 07:22 Hi, in a book I've found this about the compiler optimization "inline expansion": >>A function is more likely to be inlined by a static inliner if it is only called by one procedure, as this increases spatial locality in the instruction cache without a decrease in temporal locality.<< What I don't understand is why inlining does not decrease temporal locality in the I-cache. When I inline a function than I actually have always a loss of temporal locality: ins0 call X ins1 When X is inlined here then it is very likely that the instruction ins0 and ins1 being executed in the near future will be moved far away from each other and so possibly two I-cache accesses are required to get ins0 and ins1 while in the non-inlined function ins0,call and in1 could fit in one cache line. Do you have any idea why the book might be right with the statement about the non-decreased temporal locality? Regards, Chris
From: Ben Bacarisse on 19 Apr 2008 09:39
Stephan Ceram <linuxkaffee_@_gmx.net> writes: > Hi, > > in a book I've found this about the compiler optimization > "inline expansion": > >>>A function is more likely to be inlined by a static inliner > if it is only called by one procedure, as this increases spatial > locality in the instruction cache without a decrease in temporal > locality.<< > > What I don't understand is why inlining does not decrease temporal > locality in the I-cache. When I inline a function than I actually > have always a loss of temporal locality: > > ins0 > call X > ins1 The spacial locality is the addressing distance between executed instructions. Inlining X will increase the "spacial" locality in the instruction cache because there will be no need to jump to possibly distant X and back. The temporal locality is, presumably, the time gap between the execution of various instructions. Inlining X does not obviously increase the gap in time between executing any of the instructions involved. In fact it is likely to reduce it a bit (the gap between ins0 and ins1 is likely to be slightly shorter, rather than longer). > When X is inlined here then it is very likely that the instruction > ins0 and ins1 being executed in the near future will be moved > far away from each other But only in addressing distance, not temporal distance. > and so possibly two I-cache accesses are > required to get ins0 and ins1 while in the non-inlined function > ins0,call and in1 could fit in one cache line. Again, this is about the addressing gap between them. Of course, the non-local jump implied by the call is likely to clear the instruction cache anyway. -- Ben. |