manually unpack txid in txin #2501
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dcoutts
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Sep 29, 2021
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| TxInCompact32 :: | ||
| HS.SizeHash (CC.HASH crypto) ~ 32 => | ||
| {-# UNPACK #-} !Word64 -> -- Hash part 1/4 | ||
| {-# UNPACK #-} !Word64 -> -- Hash part 2/4 | ||
| {-# UNPACK #-} !Word64 -> -- Hash part 3/4 | ||
| {-# UNPACK #-} !Word64 -> -- Hash part 4/4 | ||
| {-# UNPACK #-} !Word64 -> -- Index |
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Yes, this is clearly 6 words, so it's a win.
| @@ -431,7 +440,10 @@ deriving newtype instance CC.Crypto crypto => FromCBOR (TxId crypto) | |||
| deriving newtype instance CC.Crypto crypto => NFData (TxId crypto) | |||
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| instance HeapWords (TxIn crypto) where | |||
| heapWords (TxInCompact txid ix) = 3 + HW.heapWordsUnpacked txid + HW.heapWordsUnpacked ix | |||
| heapWords (TxInCompact32 a _ _ _ ix) = | |||
| 6 + (4 * HW.heapWordsUnpacked a) + HW.heapWordsUnpacked ix | |||
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Are we sure that heapWordsUnpacked (a :: Word64) is zero? The answer here must be 6, so it must amount to 6 + 4 * 0 + 0. Can someone verify that the answer here does indeed come out to be simply 6?
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We have
instance HeapWords Word64 where
heapWords _ = 2
and
heapWordsUnpacked :: HeapWords a => a -> Int
heapWordsUnpacked x = heapWords x - 2
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| case HS.viewHash32 (extractHash sh) of | ||
| HS.ViewHashNot32 -> TxInCompactOther txid (fromIntegral index) | ||
| HS.ViewHash32 a b c d -> TxInCompact32 a b c d (fromIntegral index) |
Co-authored-by: Jared Corduan <jared.corduan@iohk.io>
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