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();
> }
> }
>
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