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Comment by nothrabannosir | Hacker News Reader

I always wondered: what determines the optimal block size, and how can I know?

In theory the native block size (512 bytes for most drives these days) should be the fastest, but the problem is that if you're doing such small sized I/O, you introduce a shitload of overhead for all the individual r/w calls - I guess that a huge blocksize benefits from DMA and look-ahead reads.

> native block size (512 bytes for most drives these days)

Not anymore. Advanced Format (4096-byte sector) hard drives have taken the market like a storm, and SSDs benefit even more from using larger I/O sizes (because erase sectors are way larger).

Ah, I thought it was the other way round. Thanks for the information.

Do you know a way to get an SSD's native erase sector size?

I would guess that using erase sector size as a block size won't help, because SSD controller anyway rearranges blocks all the time.

There is no simple answer to this. Stack Overflow has some hints but no definite answer.

For some reason I cannot reply to your message directly, only to a parent.

Flashbench [1] should be able to tell you what the erase-block and page size is.

[1] https://github.com/bradfa/flashbench

Isn't it: the larger the better until you face diminishing returns as with small block sizes the system call overhead becomes noticeable?

I've always used bs=4M for writing .iso or .img files to USB flash drives, as it gives me the best times. This is on Linux, OpenBSD, and macOS (using /dev/rdiskn for the latter two).

Which theory suggests that this should be the fastest?

The optimal block size is probably the the amount of data which can be transferred with one DMA operation.

For NVMe disks on Linux, you can find out this size with the nvme-cli [0] tool. Use "nvme id-ctrl" to find the Maximum Data Transfer Size (MDTS) in disk (LBA) blocks and "nvme id-ns" to find the LBA Data Size (LBADS). The value is then 2^MDTS * 2^LBADS byte.

For example, the Intel SSD 450 can transfer 32 blocks of 4096 byte per NVMe command, so you'd want a block size of 128 kiB.

[0] https://github.com/linux-nvme/nvme-cli

On FreeBSD, use diskinfo -v.

Also check out the BUGS section of the manpage :) https://www.freebsd.org/cgi/man.cgi?diskinfo

stat(2) the file, and use the value from the st_blksize member of struct stat.

Nononono, that's waaaayy to small. This will be something like 512B or 4K, meaning you'll burn the CPU on syscalls instead of doing meaningful work.

Even the 32K read/write size used by many utilities (shells, XYZsum, rsync and so on) can slow things down with modern/fast IO devices.

Today you'll want to use something like 32K to 2MB. Doesn't really matter around there.

If you're writing synchronously (which you shouldn't), then it becomes a tad more difficult to figure out for optimal performance.

Sync writes have a place, if I'm copying a disk image to a flash drive I typically want to know it's done when dd finishes so I can yank the drive without worrying if the write cache has been flushed.

At least a couple years ago, it was said one should not rip out USB sticks without ejecting them before, because their controller might do invisible maintenance work and that might lead to data corruption...

Hmm. Microsoft doesn't agree: https://i.imgur.com/JIZveQz.png

"Disables write caching […] you can disconnect the device safely without [unmounting it]"

Broadly speaking, flash GC/management journaling is the problem of the controller and is usually not the most broken part of it.

$ sync; sync; sync

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