On Sun, 2005-03-27 at 15:51 +0300, Denis Vlasenko wrote:
> > > It's impossible to be otherwise. A call requires
> > > that the return address be written to memory (the stack),
> > > using register indirection (the stack-pointer).
> >
> > and it's a so common pattern that it's optimized to death. Internally a
> > call gets transformed to 2 uops or so, one is push eip, the other is the
> > jmp (which gets then just absorbed by the "what is the next eip" logic,
> > just as a "jmp"s are 0 cycles)
>
> Arjan, you overlook the fact that kfree() contains 'if(!p) return;' too.
> call + test-and-branch can never be faster than test+and+branch
ok so for the non-null case you have
test-nbranch-call-test-nbranch
vs
call-test-nbranch
vs the null case where you get
test-branch
vs
call-test-branch
(I'm using nbranch here as a non-taken branch; it's also a conditional
branch and it has the same misprediction possibility)
in the non-null case with if you have *two* chances for the branch
predictor to go wrong. (and "wrong" can also mean "cold, eg unknown"
here) and always an extra "test-nbranch" sequence, which is probably a
cycle at least
the offset for that is the null-case-without-if where you have an extra
"call", which is also half to a whole cycle.
even in the null case it's dubious if there is gain, it depends on how
the branch predictor happens to feel that day ;)
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