Introduce a new label that includes kfree and jump to that one.

A simplified version of the semantic match that finds this problem is as
follows: (http://coccinelle.lip6.fr/)

// <smpl>
@@
identifier x;
expression E1!=0,E2,E3,E4;
statement S;
iterator I;
@@

(
if (...) { ... when != kfree(x)
               when != x = E3
               when != E3 = x
*  return ...;
 }
... when != x = E2
    when != I(...,x,...) S
if (...) { ... when != x = E4
 kfree(x); ... return ...; }
)
// </smpl>
Signed-off-by: NJulia Lawall <julia@diku.dk>
Acked-by: NAlexander Sverdlin <subaparts@yandex.ru>
Reviewed-by: NH Hartley Sweeten <hsweeten@visionengravers.com>
Acked-by: NLiam Girdwood <lrg@ti.com>
Signed-off-by: NMark Brown <broonie@opensource.wolfsonmicro.com>