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# -*- coding: UTF-8 -*-
"""Directed graph production.
This module contains the code to produce an ordered directed graph of a
bzr branch, such as we display in the tree view at the top of the bzrk
window.
"""
__copyright__ = "Copyright © 2005 Canonical Ltd."
__author__ = "Scott James Remnant <scott@ubuntu.com>"
from bzrlib.revision import Revision
from bzrlib.tsort import merge_sort
def linegraph(revisions, revisionparents, revindex):
"""Produce a directed graph of a bzr branch.
Returns a list of tuples of (revision, node, lines, parents, children).
Node is a tuple of (column, colour) with column being a zero-indexed
column number of the graph that this revision represents and colour
being a zero-indexed colour (which doesn't specify any actual colour
in particular) to draw the node in.
Lines is a list of tuples which represent lines you should draw away
from the revision, if you also need to draw lines into the revision
you should use the lines list from the previous iteration. Each
typle in the list is in the form (start, end, colour) with start and
end being zero-indexed column numbers and colour as in node.
It's up to you how to actually draw the nodes and lines (straight,
curved, kinked, etc.) and to pick the actual colours for each index.
"""
directparentcache = [None for revision in revisions]
def getdirectparent(childindex, childsparents):
"""Return the revision id of the direct parent
The direct parent is the first parent with the same committer"""
childrevision = revisions[childindex]
directparent = directparentcache[childindex]
if directparent is None:
for parentrevid in childsparents:
parentrevision = revisions[revindex[parentrevid]]
if childrevision.committer == parentrevision.committer:
directparent = parentrevid
break
#no parents have the same commiter
if directparent is None:
directparent = ""
directparentcache[childindex] = directparent
return directparent
#This will hold the lines we have not yet added to lines
#The position of the item in this list indicates the column, and it
#it may change if we need to make space for other branches.
#Each typle in the list is in the form (child index, parent revision)
#A item may be None to indicate that there is no line for a column
activelines = []
linegraph = []
lastcolor = 0
for (index, revision) in enumerate(revisions):
parents = [parent for parent in revisionparents[index]\
if parent!="null:"]
revnodecolumn = None
#This will hold a list of lines whose parent is this rev
linesforcurrentrev = []
children = []
#We should maybe should pop None's at the end of activelines.
#I'm not sure what will cost more: resizing the list, or
#have this loop ittrate more.
#Find lines that end at this rev
for (column, activeline) in enumerate(activelines):
if (activeline is not None):
(childindex, parentrevid) = activeline
if parentrevid == revision.revision_id:
linesforcurrentrev.append((childindex, parentrevid, column))
activelines[column] = None
children.append(linegraph[childindex][0].revision_id)
#The node column for this rev will be the smallest
#column for the lines that end at this rev
#The smallest column is the first one we get to.
if revnodecolumn is None:
revnodecolumn = column
#This will happen for the latest revision
if revnodecolumn is None:
revnodecolumn = 0
color = None
#Try and see if we are the same "branch" as one of our children
#If we are, use the childs color
for childrevid in children:
childindex = revindex[childrevid]
childsparents = revisionparents[childindex]
if len(children) == 1 and len(childsparents) == 1:
# one-one relationship between parent and child, same colour
#1st [1] selects the node
#2nd [1] selects the color
color = linegraph[childindex][1][1]
break
#Is the current revision the direct parent of the child?
if revision.revision_id == getdirectparent(childindex, childsparents):
color = linegraph[childindex][1][1]
break
if color is None:
color = lastcolor = lastcolor + 1
#We now have every thing (except for the lines) so we can add
#our tuple to our list.
linegraph.append((revision, (revnodecolumn, color),
[], parents, children))
#add all the line bits to the rev that the line passes
for (childindex, parentrevid, column) in linesforcurrentrev:
if index>childindex+1:
#out from the child to line
linegraph[childindex][2].append(
(linegraph[childindex][1][0], #the column of the child
column, #the column of the line
color))
#down the line
for linepartindex in range(childindex+1, index-1):
linegraph[linepartindex][2].append(
(column, #the column of the line
column, #the column of the line
color))
#in to the parent
linegraph[index-1][2].append(
(column, #the column of the line
revnodecolumn, #the column of the parent
color))
else:
#child to parent
linegraph[childindex][2].append(
(linegraph[childindex][1][0], #the column of the child
revnodecolumn, #the column of the parent
color))
for parentrevid in parents:
column = revnodecolumn
line = (index,parentrevid)
while True:
if column<len(activelines):
if activelines[column] is None:
#An empty column. Put line here
activelines[column] = line
break
else:
if activelines[column][0] == index:
#This column is allready used for a line for
#this rev, Move along.
column += 1
else:
#This column is allready used for a line for
#another rev. Insert this line at this column,
#and in the process, move all the other lines out.
activelines.insert(column, line)
break
else:
#no more columns, so add one to the end
activelines.append(line)
break
return linegraph
def same_branch(a, b):
"""Return whether we think revisions a and b are on the same branch."""
if len(a.parent_ids) == 1:
# Defacto same branch if only parent
return True
elif a.committer == b.committer:
# Same committer so may as well be
return True
else:
return False
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