The cost for scalarized operations is computed as N * (scalar operation
cost + 1 extractelement + 1 insertelement). This partially fixes
inflating the cost of scalarized operations since every operation is
scalarized and free. I don't think we want any cost asociated with
scalarization, but for now insertelement is still counted. I'm not sure
if we should pretend that insertelement is also free, or add a way
to compute a custom scalarization cost.
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- Reviewers
• tstellarAMD
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