Thanks, it's a good idea. Unfortunately the result remains, making me even more
curious.
```hs
λ> :set -XBangPatterns
λ>
λ> :set -package vector
package flags have changed, resetting and loading new packages...
λ>
λ> import Prelude
λ>
λ> import Control.Monad.ST
λ> import qualified Data.Vector.Storable as VS
λ>
λ> :{
λ|
λ| newtype SomeVector = SomeVector (VS.Vector Int)
λ|
λ| isSameVector :: SomeVector -> SomeVector -> Bool
λ| isSameVector (SomeVector !x) (SomeVector !y) =
λ| x'offset == y'offset && x'fp == y'fp
λ| where
λ| (x'fp, x'offset, _x'len) = VS.unsafeToForeignPtr x
λ| (y'fp, y'offset, _y'len) = VS.unsafeToForeignPtr y
λ| :}
λ>
λ> let !v = VS.fromList [5..200000] in isSameVector (SomeVector v) (SomeVector
v)
False
λ>
λ> let !v = SomeVector (VS.fromList [3,2,5]) in isSameVector v v
True
λ>
```
> On 2021-04-06, at 20:00, Henning Thielemann <[email protected]>
> wrote:
>
>
> On Tue, 6 Apr 2021, YueCompl via Haskell-Cafe wrote:
>
>> In an attempt to determine whether two immutable vectors can be treated as
>> the same one to enable specific optimizations for that case, I tried to use
>> ST to determine their respective backing foreign ptrs for comparison. But
>> appears it can be copied when wrapped in a newtype, I wonder why it is the
>> case, and how to avoid the copy?
>
> You compare the ForeignPtrs of the mutable vectors. What about comparing the
> ForeignPtrs of the original immutable vectors?
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