Re: [PATCH 1/2] pci: endpoint: Free func_name after last usage

[Lorenzo Pieralisi <lorenzo.pieralisi@arm.com>]

On Wed, Feb 21, 2018 at 01:47:06PM +0100, Rolf Evers-Fischer wrote:
> From: Rolf Evers-Fischer <rolf.evers.fischer@aptiv.com>
> 
> This commit decreases the number of jump labels and ensures
> that the next commit doesn't increase the number of occurrences
> of 'kfree(func_name)'.
> 
> Change-Id: I0d1b6fd652395b85f82b11c43bf9b7db512854d1
> Signed-off-by: Rolf Evers-Fischer <rolf.evers.fischer@aptiv.com>
> Signed-off-by: Rolf Evers-Fischer <embedded24@evers-fischer.de>
> ---
>  drivers/pci/endpoint/pci-epf-core.c | 7 ++-----
>  1 file changed, 2 insertions(+), 5 deletions(-)
> 
> diff --git a/drivers/pci/endpoint/pci-epf-core.c b/drivers/pci/endpoint/pci-epf-core.c
> index 766ce1dca2ec..23d0e128d1a5 100644
> --- a/drivers/pci/endpoint/pci-epf-core.c
> +++ b/drivers/pci/endpoint/pci-epf-core.c
> @@ -220,9 +220,10 @@ struct pci_epf *pci_epf_create(const char *name)
>  	*buf = '\0';
>  
>  	epf->name = kstrdup(func_name, GFP_KERNEL);
> +	kfree(func_name);

I am certainly missing something but what if we reworked the code
and just:

kstrdup(name, GFP_KERNEL);

once instead of allocating another local copy (that we then have to
free) ?

Reworded: why

epf->name = func_name;

would not work ?

Thanks,
Lorenzo

>  	if (!epf->name) {
>  		ret = -ENOMEM;
> -		goto free_func_name;
> +		goto free_epf;
>  	}
>  
>  	dev = &epf->dev;
> @@ -238,16 +239,12 @@ struct pci_epf *pci_epf_create(const char *name)
>  	if (ret)
>  		goto put_dev;
>  
> -	kfree(func_name);
>  	return epf;
>  
>  put_dev:
>  	put_device(dev);
>  	kfree(epf->name);
>  
> -free_func_name:
> -	kfree(func_name);
> -
>  free_epf:
>  	kfree(epf);
>  
> -- 
> 2.16.2
>