On Wed, 9 Mar 2011, Dan McGee wrote:
> On Wed, Mar 9, 2011 at 10:09 AM, Julia Lawall <[email protected]> wrote:
> > The solution is not to actually perform any transformations on the
> > subexpressions a and b. The example works fine with the following
> > semantic patch:
> >
> > @ oper_add @
> > expression a, b;
> > @@
> > - oper_add(
> > a
> > - ,
> > + +
> > b
> > - )
> >
> > The only small unfortunate bit is that the pretty printing is not very
> > nice. It doesn't realize that + is a binary operator, and so there is no
> > space before it, although there is a space after it. It should be
> > possible to fix this.
>
> Thanks for the hint here- I ended up finding something that did work
> closer to expected and formats it relatively nicely as it seems to
> recognize the binary operator:
> @@
> expression a, b;
> @@
> - oper_add(a,
> + a +
> b
> - )
>
> One more question. I'm looking now at a follow-on transform to the
> above to simplify things more, and this works for the simple case, but
> not the follow-on cases.
> a = a + 1; --> a+= 1;
> b = b + a + 1 --> does not get transformed.
>
> expression a, b;
> @@
> - a = ((a) + (b));
> + a += b;
This is a bit more complicated. The problem is the associativity of +. +
associates to the left, and you want it to associate to the right. If you
write a + ..., ie with ... as the other argument, then it will consider
any sum with a at the top level. This works for various associative
operators (eg || and &&). Butthe problem is that you can't use ...
because you need b to construct the new code. And if you organize things
such that b doesn't have to be in the + code, ie it could be expressed as
..., then coccinelle is not able to verify that this isomorphism for +
and ... is safe to apply.
A solution is to identify the case first and then make the transformation
in a separate rule:
@r@
expression a;
position p;
@@
a =@p a + ...;
@@
expression a, b;
position r.p;
@@
a
- =@p
+ +=
<+...
- a + b
+ b
...+>;
This works for the first two ofteh following functions, but not for the
third one, because then there are two matches that express different
transformations:
int main () {
b = b + a + 1;
}
int main () {
b = 1 + a + b;
}
int main () {
b = b + a + b;
}
julia_______________________________________________
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