Hi Rajendra, On Tue, Nov 8, 2016 at 9:23 AM, Rajendra Nayak <[email protected]> wrote: > With clk_hw_register() API we hide the struct clk from the caller > and return an int error code instead, so the caller (clk provider) > is not expected to use hw->clk on return.
That's correct, in case of failure. > Free the memory, and mark hw->clk as NULL before returning. > > Signed-off-by: Rajendra Nayak <[email protected]> > --- > drivers/clk/clk.c | 10 +++++++++- > 1 file changed, 9 insertions(+), 1 deletion(-) > > diff --git a/drivers/clk/clk.c b/drivers/clk/clk.c > index 0fb39fe..f81e4aa 100644 > --- a/drivers/clk/clk.c > +++ b/drivers/clk/clk.c > @@ -2628,7 +2628,15 @@ struct clk *clk_register(struct device *dev, struct > clk_hw *hw) > */ > int clk_hw_register(struct device *dev, struct clk_hw *hw) > { > - return PTR_ERR_OR_ZERO(clk_register(dev, hw)); > + struct clk *c; > + > + c = clk_register(dev, hw); > + if (IS_ERR(c)) > + return PTR_ERR(c); > + > + __clk_free_clk(c); > + hw->clk = NULL; This is the success path, not the failure path (on failure, clk_register() has already freed the struct clk). Why do you free the struct clk in case of success? What am I missing? > + return 0; > } > EXPORT_SYMBOL_GPL(clk_hw_register); Gr{oetje,eeting}s, Geert -- Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- [email protected] In personal conversations with technical people, I call myself a hacker. But when I'm talking to journalists I just say "programmer" or something like that. -- Linus Torvalds
