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by tmoertel
4749 days ago
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They're not orthogonal; they're two directed graphs with the same vertices and different edges. If the dependency graph G = (V, E) has a vertex for every software project and an edge x -> y iff downstream project y depends on upstream project x, then the change-flow network is the graph C = (V, F), where there is an edge x -> y in F iff there is a downstream path between x and y in G and also y requires an update and re-release when x changes (e.g., because it bundles a copy of x in its releases). So if there is a change to project x, for it to flow to all affected dependents, you must update all downstream neighbors of x in the change-flow network C. For example, consider the following dependency graph, in which library L is used by downstream library L2, and L2 by project P: L -> L2 -> P
If none of the projects bundle their upstream dependencies in their own releases, then the corresponding change-flow network has no edges, and updating any project requires only re-releasing its own package to satisfy all dependencies: L
L2
P
But if L2 bundles a copy of L, and P bundles a copy of L2, then the corresponding network looks like this: L -> L2
L -> P
L2 -> P
P
A change to L requires re-releasing not only L but also L2, and P. A change to L2 requires re-releasing L2 and also P.Does that make more sense now? |
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If P statically links to its own version of L2, then L2 is just a part of P. The fact that there may be a dynamically linked version of L2 elsewhere on the system is irrelevant.
Consider:
If the authors of L2 release a new version that P and Q are happy with, but creates an extremely subtle segfault condition in R, then what?The packager could just wait to release the upgrade to L2 until all downstream packages have compatible releases.
The packager could backport a subset of the L2 patches that is still compatible with R (Redhat does this a lot).
The packager could silently curse the author of R for not statically linking the necessary frozen-in-time version of L2 and thus bypassing this problem entirely.