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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T12:03:48+00:00 2026-05-13T12:03:48+00:00

I have executed a query and included the Actual Execution Plan. There is one

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I have executed a query and included the Actual Execution Plan. There is one Hash Match that is of interest to me because it’s subtree uses a Index Scan instead of an index seek. When I mouse over this Hash Match there is a section called “Probe Residual”. I had assumed that this is whatever values I am joining on. Am I correct here or is there a better explanation of what that means?

The second question I had is regarding the indexes it uses. In my example I am pretty sure this particular join is joining on two columns. The index that it is Scanning has both of these columns in it as well as another column that is not used in the join. I was under the impression that this would result in an Index Seek rather than a Scan. Am I mistaken on this?

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  1. Editorial Team
    Editorial Team
    2026-05-13T12:03:48+00:00Added an answer on May 13, 2026 at 12:03 pm

    A Hash Join will generally (always?) use a scan or at least a range scan. A hash join works by scanning both left and right join tables (or a range in the tables) and building an in-memory hash table that contains all values ‘seen’ by the scans.

    What happened in your case is this: the QO noticed that it can obtain all the values of a column C from a non-clustered index that happens to contain this column (as a key or as an included column). Being a non-clustered index is probably fairly narrow, so the total amount of IO to scan the entire non-clustered index is not exaggerate. The QO also considered that the system has enough RAM to store a hash table in memory. When compared the cost of this query (a scan of a non-clustered index end-to-end for, say, 10000 pages) with the cost of a nested loop that used seeks (say 5000 probes at 2-3 pages each) the scan won as requiring less IO. Of course, is largely speculation on my part, but I’m trying to present the case from the QO point of view, and the plan is likely optimal.

    Factors that contributed to this particular plan choice would be:

    • a large number of estimated candidates on the right side of the join
    • availability of the join column in a narrow non-clustered index for the left side
    • plenty of RAM

    For a large estimate of the number of candidates, a better choice than the hash join is only the merge-join, and that one requires the input to be presorted. If both the left side can offer an access path that guarantees an order on the joined column and the right side has a similar possibility then you may end up with the merge join, which is the fastest join.

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