On 2016/5/26 21:13, Rob Herring wrote:
> On Thu, May 26, 2016 at 10:43:58AM +0800, Zhen Lei wrote:
>> For a normal memory@ devicetree node, its reg property can contains more
>> memory blocks.
>>
>> Because we don't known how many memory blocks maybe contained, so we try
>> from index=0, increase 1 until error returned(the end).
>>
>> Signed-off-by: Zhen Lei <thunder.leiz...@huawei.com>
>> ---
>>  drivers/of/of_numa.c | 30 ++++++++++++++++++++----------
>>  1 file changed, 20 insertions(+), 10 deletions(-)
>>
>> diff --git a/drivers/of/of_numa.c b/drivers/of/of_numa.c
>> index 21d831f..2c5f249 100644
>> --- a/drivers/of/of_numa.c
>> +++ b/drivers/of/of_numa.c
>> @@ -63,7 +63,7 @@ static int __init of_numa_parse_memory_nodes(void)
>>      struct device_node *np = NULL;
>>      struct resource rsrc;
>>      u32 nid;
>> -    int r = 0;
>> +    int i, r = 0;
>>
>>      for (;;) {
>>              np = of_find_node_by_type(np, "memory");
>> @@ -82,17 +82,27 @@ static int __init of_numa_parse_memory_nodes(void)
>>                      /* some other error */
>>                      break;
>>
>> -            r = of_address_to_resource(np, 0, &rsrc);
>> -            if (r) {
>> -                    pr_err("NUMA: bad reg property in memory node\n");
>> -                    break;
>> +            for (i = 0; ; i++) {
>> +                    r = of_address_to_resource(np, i, &rsrc);
>> +                    if (r) {
>> +                            /* reached the end of of_address */
>> +                            if (i > 0) {
>> +                                    r = 0;
>> +                                    break;
>> +                            }
>> +
>> +                            pr_err("NUMA: bad reg property in memory 
>> node\n");
>> +                            goto finished;
>> +                    }
>> +
>> +                    r = numa_add_memblk(nid, rsrc.start,
>> +                                        rsrc.end - rsrc.start + 1);
>> +                    if (r)
>> +                            goto finished;
>>              }
>> -
>> -            r = numa_add_memblk(nid, rsrc.start,
>> -                                rsrc.end - rsrc.start + 1);
>> -            if (r)
>> -                    break;
>>      }
>> +
>> +finished:
>>      of_node_put(np);
> 
> This function can be simplified down to:
> 
>       for_each_node_by_type(np, "memory") {
OK, That's good.

>               r = of_property_read_u32(np, "numa-node-id", &nid);
>               if (r == -EINVAL)
>                       /*
>                        * property doesn't exist if -EINVAL, continue
>                        * looking for more memory nodes with
>                        * "numa-node-id" property
>                        */
>                       continue;
Hi, everybody:
    If some "memory" node contains "numa-node-id", but some others missed. Can 
we simply ignored it?
I think we should break out too, and faking to only have node0.

>               else if (r)
>                       /* some other error */
>                       break;
> 
>               r = of_address_to_resource(np, 0, &rsrc);
>               for (i = 0; !r; i++, r = of_address_to_resource(np, i, 

But r(non-zero) is just break this loop, the original is break the outer for 
(;;) loop

How about as below?

        for_each_node_by_type(np, "memory") {
                ... ...

                for (i = 0; !of_address_to_resource(np, i, &rsrc); i++) {
                        r = numa_add_memblk(nid, rsrc.start,
                                            rsrc.end - rsrc.start + 1);
                        if (r)
                                goto finished;
                }

                if (!i)
                        pr_err("NUMA: bad reg property in memory node\n");
        }

finished:
        

> &rsrc)) {
>                       r = numa_add_memblk(nid, rsrc.start,
>                                   rsrc.end - rsrc.start + 1);
>               }
>       }
>       of_node_put(np);
> 
>       return r;
> 
> 
> Perhaps with a "if (!i && r) pr_err()" for an error message at the end.
> 
> Rob
> 
> .
> 

Reply via email to