[SOSTests] Avoid double counting GCRoots #5595
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If anyone has a more appropriate fix, I'm all ears. I'm not familiar with SOS tests/heaps logic. |
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This change looks good to me. The idea of a 'unique' root depends on how you think about objects being rooted in multiple heaps. Do we care about 'unique' roots per heap or per object. This approach makes sense to me as counting the same object as multiple roots is confusing. However, I don't have much experience with the GC behavior either. If we don't want to change the commands behavior, we can update the test to verify that Is this behavior change okay @hoyosjs ? |
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From some more digging of how GCRoots works, my current mental model is that That way, in all paths outside the one calling PrintOlderGenerationRoots, theres a claim its all unique roots. if (gen < 0 || gen > 1)
{
// If not gen0 or gen1, treat it as a normal !gcroot
if (NoStacks)
{
count = PrintNonStackRoots(gcroot, limit); <- Unique
}
else
{
count = PrintAllRoots(gcroot, limit); <- Unique
}
}
======================
else
{
count = PrintOlderGenerationRoots(gcroot, gen, limit);
count += PrintNonStackRoots(gcroot, limit);
}
======================
}
else if (NoStacks)
{
count = PrintNonStackRoots(gcroot, limit); <- Unique
}
else
{
count = PrintAllRoots(gcroot, limit); <- Unique
}
Console.WriteLine($"Found {count:n0} unique roots.");On the other hand, I'm not caught up with how the test validates this runtime change dotnet/runtime#119396, but if just checking |
3 participants
On Win-x86,
FindRootsOlderGenerationfailed with double counting the same GC Root. My guess is that the same GCRoot was found on multiple subheaps, so adding deduplication logic during counting would account for that.Example failing test