Comments (4)
Feel free to pm me if you need anything to debug.
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You can't encode every Go struct. The struct needs to have encoding methods or contain public fields.
Tree is not meant to be encoded. There is no need. If you want to save a tree, just save its elements.
When you load it, just reinsert the elements in the tree.
from gollrb.
Thanks for the reply. I was simply playing with gollrb. It seems to me that it would be faster to read from disk in gob encoded than reading and re-inserting elements in the tree. Lets say that you want to save the current state of the tree so you can process it later. Also, wouldn't the rpc package fail if you try to send this since they rely on gob?. Thanks again.
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It is marginally faster to save the tree structure, rather than to save the elements and reinsert them.
The new Root() method exposes the root element and you can serialize it with whatever (gob, json, etc.)
When de-serializing, use SetRoot.
Note that the comparison function cannot be serialized.
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Related Issues (16)
- Sedgewick's misnaming
- IterRange can never include the last (Max) element HOT 3
- Example Code does not compile HOT 1
- top down 2-3-4 mode does not correctly delete HOT 3
- Code assumes root node has been set HOT 1
- I've forked GoLLRB to be quasi-immutable.
- Still active? HOT 4
- Particular data sequence causes inconsistency & panic in llrb
- Anybody who wants to follow this project can reference my fork HOT 4
- please tag formal releases
- please clarify license and copyright of files in "doc" HOT 1
- panic at GoLLRB/llrb/llrb.go:426
- LLRB merge/union algorithm
- Example docs became invalid
- Panic in Delete HOT 1
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