Jules Bean wrote:
I don't think it does, actually. You can imagine a compiler which has
access to not *only* the .so files, but also the haskell source.
Therefore it can still unroll (from the source), but it can choose to
link to an exported symbol if unrolling isn't worth it.
But that's not dynamic linking... Imagine a bug in version X of your lib,
simply using version X+1 with your already compiled program won't fix that
bug. Again, this is just like C++.
Cheers,
S.
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