Re: [PATCH] btrfs-progs: print-tree: Enehance uuid item print

[Qu Wenruo <quwenruo.btrfs@gmx.com>]

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On 2017年10月31日 15:15, Nikolay Borisov wrote:
> 
> 
> On 31.10.2017 06:03, Qu Wenruo wrote:
>> For key type BTRFS_UUID_KEY_SUBVOL or BTRFS_UUID_KEY_RECEIVED_SUBVOL the
>> key objectid and key offset are just half of the UUID.
>>
>> However we just print the key as %llu, which is converted from little
>> endian, not byte order for UUID, nor the traditional 36 bytes human
>> readable uuid format.
>>
>> Although true engineer can easily convert it in their brain, but to
>> make it easier for search, output the result UUID using the 36 chars format.
>>
>> Cc: Misono Tomohiro <misono.tomohiro@jp.fujitsu.com>
>> Signed-off-by: Qu Wenruo <wqu@suse.com>
>> ---
>> Inspired by UUID related work from Misono.
>> ---
>>  print-tree.c | 17 ++++++++++++++---
>>  1 file changed, 14 insertions(+), 3 deletions(-)
>>
>> diff --git a/print-tree.c b/print-tree.c
>> index 3c585e31f1fc..687f871db302 100644
>> --- a/print-tree.c
>> +++ b/print-tree.c
>> @@ -803,14 +803,25 @@ void btrfs_print_key(struct btrfs_disk_key *disk_key)
>>  	}
>>  }
>>  
>> -static void print_uuid_item(struct extent_buffer *l, unsigned long offset,
>> -			    u32 item_size)
>> +static void print_uuid_item(struct extent_buffer *l, int slot,
>> +			    unsigned long offset, u32 item_size)
>>  {
>> +	struct btrfs_key key;
>> +	char uuid_str[BTRFS_UUID_UNPARSED_SIZE];
>> +	u8 uuid[BTRFS_UUID_SIZE];
>> +
>> +	/* Reassemble the uuid from key.objecitd and key.offset */
>> +	btrfs_item_key_to_cpu(l, &key, slot);
>> +	put_unaligned_le64(key.objectid, uuid);
>> +	put_unaligned_le64(key.offset, uuid + sizeof(u64));
> 
> I don't think this will work on a BE system. Because
> btrfs_item_key_to_cpu take the LE representation on-disk and turns it
> into a cpu representation which might very well be BE. And then you
> essentially reverse it by using put_unaligned_le64 for x86 it works fine
> due to it being a LE system.

I know this can be tricky, let's assume the following case:

UUID: 0x0123456789abcdef0123456789abcdef (byte order, no endian)
              Low bit          high bit
Key objectid: 0x0123456789abcdef (LE on-disk)

              0xefcdab8967452301 (u64)  <- CPU key.objectid

              Low bit         hight bit
key.offset:   0x123456789abcdef (LE on-disk)
              0xefcdab8967452301 (u64)  <- CPU key.offset

put_unaligned_le64 will convert CPU key.objectid/offset to LE on-disk
again, so we get
              Low bit                  high bit
              0x01    23456789abcd     ef (LE on-disk)
              uuid[0]      ...         uuid[7]

And that's what we need.

We did the LE->native and native->LE, so the result is not changed at all.
And we just need byte order, so the result is correct.

Just like what kernel did.

Thanks,
Qu
> 
> 
>> +	uuid_unparse(uuid, uuid_str);
>> +
>>  	if (item_size & (sizeof(u64) - 1)) {
>>  		printf("btrfs: uuid item with illegal size %lu!\n",
>>  		       (unsigned long)item_size);
>>  		return;
>>  	}
>> +	printf("\t\tuuid %s\n", uuid_str);
>>  	while (item_size) {
>>  		__le64 subvol_id;
>>  
>> @@ -1297,7 +1308,7 @@ void btrfs_print_leaf(struct btrfs_root *root, struct extent_buffer *eb)
>>  			break;
>>  		case BTRFS_UUID_KEY_SUBVOL:
>>  		case BTRFS_UUID_KEY_RECEIVED_SUBVOL:
>> -			print_uuid_item(eb, btrfs_item_ptr_offset(eb, i),
>> +			print_uuid_item(eb, i, btrfs_item_ptr_offset(eb, i),
>>  					btrfs_item_size_nr(eb, i));
>>  			break;
>>  		case BTRFS_STRING_ITEM_KEY: {
>>
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