`On Thu, Mar 08, 2012 at 10:22:26AM -0800, Arun Sharma wrote:> On Thu, Mar 08, 2012 at 04:39:36PM +0100, Frederic Weisbecker wrote:> > > > I don't yet understand the point of this.> > > > Imagine those three hists:> > > > a -> b -> c> > a -> b -> d> > a-> e -> f> > > > The fractal inverted mode (-G) will report this:> > > > a--> >   |> >   ----- b> >   |     |> >   |     -----c> >   |     |> >   |     -----d> >   |> >   ----- e> >         |> >         -----f> > > > Please see the test program attached. It has only two paths from main()> to c(). But for this discussion, imagine a callgraph with 10 different> paths.Ok that makes sense.Thanks.> > With -G, c() appears 10 times in the callgraph and the user is required> to manually sum up the samples to realize that the callgraph under c()> is very expensive.> > With -s inclusive, c() will show up at the very top after main().> >  -Arun> > #include <stdio.h>> > int sum = 0;> > #define LOOP(n)                                         \>         {                                               \>                 int j;                                  \>                 for (j = 0; j < 10000; j++) {           \>                         sum += j;                       \>                 }                                       \>         }> > int f()> {>         LOOP(100);> }> > int d()> {>         LOOP(100);>         f();> }> > int e()> {> 	LOOP(100);> 	f();> }> > int c()> {> 	LOOP(100);> 	d();> 	LOOP(100);> 	e();> }> > int b() > {> 	LOOP(70);> 	c();> }> > int a() > {> 	LOOP(30);> 	c();> }> > int main()> {> 	int i;> 	for (i = 0; i < 10000; i++) {> 		a();> 		b();> 	}> }> `