I see, thanks! It's a relief, that huge overhead (as I wrongly perceived) really made me uncomfortable.

On 2020-07-28, at 00:21, Oleg Grenrus wrote:

TL;DR bits, not bytes.

It meant to say 320 bit.

4 * 64 (Each Word32 is stored as Word64) + one 64bit header.

5 * 64 = 320.

It could be just 3 * 64 = 192.

- Oleg

On 27.7.2020 9.58, YueCompl via Haskell-Cafe wrote:

...

Hello Cafe,

I'm about to introduce UUID into my code, and see https://github.com/haskell-hvr/uuid/issues/24 https://github.com/haskell-hvr/uuid/issues/24 stating:

...

Currently, UUID is represented as data UUID = UUID {-# UNPACK #-} !Word32 {-# UNPACK #-} !Word32 {-# UNPACK #-} !Word32 {-# UNPACK #-} !Word32 However, this suboptimal for 64bit archs (where GHC currently stores this a 320-byte Heap object); ...

According to https://wiki.haskell.org/GHC/Memory_Footprint https://wiki.haskell.org/GHC/Memory_Footprint I can understand each evaluated `Word32` on 64-bit hardware can take 2 words - 16 bytes, and given they are unpacked and strict, I think one whole UUID record should just take 64 bytes plus a few words, which is far less than 320 bytes. So how comes the 320 bytes?

Thanks with regards, Compl

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