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I'm partitioning a disk for FreeBSD installation.

I create my partitions/slices:

gpart create -s mbr ada0
gpart add -t freebsd -s 1G ada0
gpart add -t freebsd ada0

gpart create -s bsd ada0s1

Now I get the following message:

# gpart show ada0
=>      63  488397105    ada0  MBR  (233G)
        63    2097152    1     freebsd  (1.0G)
   2097215  486299953    2     freebsd  (232G)

Which to me is perfectly fine. Then I saw the following tip in the handbook:

Proper sector alignment provides the best performance, and making partition sizes even multiples of 4K bytes helps to ensure alignment on drives with either 512-byte or 4K-byte sectors. Generally, using partition sizes that are even multiples of 1M or 1G is the easiest way to make sure every partition starts at an even multiple of 4K.

Since I've used a 1G initial partition I should be aligned at the 4K boundary?

I ask this because when I use the -a 4K flag in adding my partitions I get this extra line before my first partition instead:

=>      63  488397105    ada0  MBR  (233G)
        63          1        - free -  (512B)
        ...

Which to an inexperienced person as myself, looks like its aligning itself to a block.

This leads me to my two questions:

  1. What is this 512b block doing here?
  2. Am I losing any performance by not having this free area and going with my original plan?
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If your disk is SSD, as I think it is, it simply makes no difference. The alignment in spindle disks is a different thing, though not just as you might expect, since the kernel is not actually writing directly to the disk surface but rather to the presentation provided by the disk's software.

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