Re: [PATCH 3/3] tracing: Rewrite filter logic to be simpler and faster

[Steven Rostedt <rostedt@goodmis.org>]

On Mon, 12 Mar 2018 13:42:46 +0100
Jiri Olsa <jolsa@redhat.com> wrote:

> On Fri, Mar 09, 2018 at 09:34:45PM -0500, Steven Rostedt wrote:
> 
> SNIP
> 
> >  
> > -/* If not of not match is equal to not of not, then it is a match */
> > +/*
> > + * Without going into a formal proof, this explains the method that is used in
> > + * parsing the logical expressions.
> > + *
> > + * For example, if we have: "a && !(!b || (c && g)) || d || e && !f"
> > + * The first pass will convert it into the following program:
> > + *
> > + * n1: r=a;       l1: if (!r) goto l4;
> > + * n2: r=b;       l2: if (!r) goto l4;  
> 
> got stuck in here.. should that be 'goto l5' ?

Nope, this is correct. In fact, my user space implementation of the
code is what generated this output.

 !(!b || (c && g)) is the same as (b && (!c || !g)), so we can rewrite
 the above as:

 a && b && (!c || !g) || d || e && !f

Which makes it more obvious why it is goto l4.

And since we have:

 n1: r=a;	l1: if (!r) goto l4;
 n2: r=b;	l2: if (!r) goto l4;
 n3: r=c; r=!r;	l3: if (r) goto l4;
 n4: r=g; r=!r;	l4: if (r) goto l5;
 n5; r=d;	l5: if (r) goto T;

If a is true, if (!r) is false so we continue to n2.
If b is false, then if (!r) is true, so we take the branch. Let's see
what that does:

 l4: if (r) goto l5;

But since we jumped because r is false, then this if statement is
guaranteed to be false, and we continue to n5. Then we need to test d.

 a && b && ... || d

You can see how that is correct. If a is true and then be is false,
then we ignore the rest of the && statement and jump to testing the
other side of || which is d.

-- Steve

> 
> jirka
> 
> > + * n3: r=c; r=!r; l3: if (r) goto l4;
> > + * n4: r=g; r=!r; l4: if (r) goto l5;
> > + * n5: r=d;       l5: if (r) goto T
> > + * n6: r=e;       l6: if (!r) goto l7;
> > + * n7: r=f; r=!r; l7: if (!r) goto F
> > + * T: return TRUE
> > + * F: return FALSE  
> 
> jirka