Fixed deadlock with SetNodeInfo #188
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When the SetNodeInfo nodes n and oldNode are the same pointer there is a deadlock because the SetNodeInfo holds node n's mutex and defers the release. However, in the SetNodeInfo it calls GetInfo on the oldNode and GetInfo waits for the mutex to be released by SetNodeInfo.
bbist
reviewed
Apr 16, 2021
| @@ -595,6 +595,9 @@ func (n *Node) getTime(uri string) float64 { | |||
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| func (n *Node) SetNodeInfo(oldNode *Node) { | |||
| // Copy timers and size to new map from oldNode | |||
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Looks good.
Maybe move this comment below the equality check.
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Premise
When the SetNodeInfo's nodes n and oldNode are the same pointer there is a deadlock because the SetNodeInfo holds node n's mutex and defers the release. However, in the SetNodeInfo it calls GetInfo on the oldNode and GetInfo waits for the mutex to be released by SetNodeInfo.
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