Messages in this thread | | | From | Madper Xie <> | Subject | Re: [PATCH] Make efi-pstore return a unique id | Date | Wed, 20 Nov 2013 17:29:12 +0800 |
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richard@nod.at writes:
> Am 01.11.2013 20:22, schrieb Seiji Aguchi: >>> >>> Agreed. I liked your ((timestamp * 100 + part) * 100 + count function much >>> more than this. >> >> I was worried that the part and count could be more than 100. >> If it happens, the id may not be unique... >> >> But, currently, size of nvram storage is limited, so it is a corner case. >> I respect your opinion. > > What about feeding the bytes of all three integers into a > non-cryptographic hash function?
Then will lost the sequence of our log. We will get lots of entries like "dmesg-efi-`unique but meaningless number here`" in pstore fs. Who will know which file is the latest record? A possible way is sort them by created time. But pstore splits large messages into many parts. So they will have the same created-time. Like the following case: [root@dhcp-13-41 ~]# ls -rtl /dev/pstore/ total 0 -r--r--r--. 1 root root 930 Nov 15 08:42 dmesg-efi-9 -r--r--r--. 1 root root 1017 Nov 15 08:42 dmesg-efi-8 -r--r--r--. 1 root root 993 Nov 15 08:42 dmesg-efi-7 -r--r--r--. 1 root root 984 Nov 15 08:42 dmesg-efi-6 -r--r--r--. 1 root root 1008 Nov 15 08:42 dmesg-efi-5 -r--r--r--. 1 root root 909 Nov 15 08:42 dmesg-efi-10 -r--r--r--. 1 root root 1003 Nov 15 08:42 dmesg-efi-11 -r--r--r--. 1 root root 980 Nov 18 00:41 dmesg-efi-4 -r--r--r--. 1 root root 990 Nov 18 00:41 dmesg-efi-3 -r--r--r--. 1 root root 966 Nov 18 00:41 dmesg-efi-2 -r--r--r--. 1 root root 1010 Nov 18 00:41 dmesg-efi-1
or more intuitive: ls /sys/firmware/efi/efivars/ | grep -i "dump" | cut -d'-' -f5 | sort |wc -l 103 ls /sys/firmware/efi/efivars/ | grep -i "dump" | cut -d'-' -f5 | sort |uniq |wc -l 26
So if we using a hashed unique number for uniqueness, we will lose the sequence. We must sort them manually.
And another side, the combin of timestamp, count and part is unique. Why we generate a unique number from a unique number? if you think "making a string from three ints and then a parse it to a int again" is odd, i'd like to use ((timestamp * 100 + part) * 100 + count.
> Using this way you get a cheap unique id. > > Thanks, > //richard
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