| 1 |
On Sat, Aug 25 2012, Frank Steinmetzger wrote: |
| 2 |
|
| 3 |
> On Sun, Aug 26, 2012 at 12:22:47AM +0200, Volker Armin Hemmann wrote: |
| 4 |
> |
| 5 |
>> > > The size of an erasable block of SSDs is even larger, usually 512K, it |
| 6 |
>> > > would be best to align to that, too. A partition offset of 512K or 1M |
| 7 |
>> > > would avoid this. |
| 8 |
>> > |
| 9 |
>> > Unless the filesystem knows this and starts bigger files at those 512 k |
| 10 |
>> > boundaries (so really only one erase cycle is needed for files <=512 k), |
| 11 |
>> > isn't this fairly superfluous? |
| 12 |
>> |
| 13 |
>> no, if you misalign, a lot of 4k blocks might span into two erase blocks. |
| 14 |
>> Which is bad. If you align correctly, you will never cross them unnecessary, |
| 15 |
>> sparing your SSD some unnecessary writes and improving overall performance. |
| 16 |
> |
| 17 |
> I don’t quite follow. If you align to 4k, then you are also aligned to 512k, |
| 18 |
> because 512 % 4 = 0. |
| 19 |
|
| 20 |
Backwards. Consider starting at 8K. This is a multiple of 4K but is |
| 21 |
not a multiple of 512K. |
| 22 |
|
| 23 |
If you align to 512k than you are also aligned to 4k because 512 % 4 = |
| 24 |
0. |
| 25 |
|
| 26 |
allan |
Archives