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Differential D6578

copies: document how 'copies' dict instances are reused
ClosedPublic

Authored by martinvonz on Jun 26 2019, 8:20 AM.

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rHG898b36f74f75: copies: document how 'copies' dict instances are reused
Summary

We avoid copying these instances as much as we can, so it's not
obvious what's safe to do with them. This patch tries to explain what
is safe and what is not.

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Repository
rHG Mercurial
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Event Timeline

martinvonz created this revision.Jun 26 2019, 8:20 AM
martinvonz added a commit: rHG898b36f74f75: copies: document how 'copies' dict instances are reused.Jun 26 2019, 7:33 PM
This revision was not accepted when it landed; it landed in state Needs Review.
Closed by commit rHG898b36f74f75: copies: document how 'copies' dict instances are reused (authored by martinvonz). · Explain Why
This revision was automatically updated to reflect the committed changes.

Revision Contents
Changeset List

PathPackages
Mmercurial/copies.py (4 lines)

Diff 15667

mercurial/copies.py

children[p] = [r] children[p] = [r]
else: else:
children[p].append(r) children[p].append(r)
roots = set(children) - set(missingrevs) roots = set(children) - set(missingrevs)
# 'work' contains 3-tuples of a (revision number, parent number, copies). # 'work' contains 3-tuples of a (revision number, parent number, copies).
# The parent number is only used for knowing which parent the copies dict # The parent number is only used for knowing which parent the copies dict
# came from. # came from.
# NOTE: To reduce costly copying the 'copies' dicts, we reuse the same
# instance for *one* of the child nodes (the last one). Once an instance
# has been put on the queue, it is thus no longer safe to modify it.
# Conversely, it *is* safe to modify an instance popped off the queue.
work = [(r, 1, {}) for r in roots] work = [(r, 1, {}) for r in roots]
heapq.heapify(work) heapq.heapify(work)
alwaysmatch = match.always() alwaysmatch = match.always()
while work: while work:
r, i1, copies = heapq.heappop(work) r, i1, copies = heapq.heappop(work)
if work and work[0][0] == r: if work and work[0][0] == r:
# We are tracing copies from both parents # We are tracing copies from both parents
r, i2, copies2 = heapq.heappop(work) r, i2, copies2 = heapq.heappop(work)