upstream/mercurial-mirror Files · mercurial/dicthelpers.py

test-pathencode: randomize length of each path component...

test-pathencode: randomize length of each path component This makes it possible for long and short components to exist in the same path. This also makes shorter path components more likely. For randint(1, randint(1, n)), the likelihood that one sees a number k (1 <= k <= n) is 1/n * (\sum_{k=i}^n 1/i). This decreases with k, much like in the real world where shorter paths are more common than longer ones. The previous fix and this one together cause issue3958 to be detected by this test with reasonable frequency. When this test was run 100 times in a loop, the issue was detected 30 of those times.

Siddharth Agarwal - - Load All Authors

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      # dicthelpers.py - helper routines for Python dicts

      #

      # Copyright 2013 Facebook

      #

      # This software may be used and distributed according to the terms of the

      # GNU General Public License version 2 or any later version.

      def diff(d1, d2, default=None):

          '''Return all key-value pairs that are different between d1 and d2.

          This includes keys that are present in one dict but not the other, and

          keys whose values are different. The return value is a dict with values

          being pairs of values from d1 and d2 respectively, and missing values

          treated as default, so if a value is missing from one dict and the same as

          default in the other, it will not be returned.'''

          res = {}

          if d1 is d2:

              # same dict, so diff is empty

              return res

          for k1, v1 in d1.iteritems():

              v2 = d2.get(k1, default)

              if v1 != v2:

                  res[k1] = (v1, v2)

          for k2 in d2:

              if k2 not in d1:

                  v2 = d2[k2]

                  if v2 != default:

                      res[k2] = (default, v2)

          return res

      def join(d1, d2, default=None):

          '''Return all key-value pairs from both d1 and d2.

          This is akin to an outer join in relational algebra. The return value is a

          dict with values being pairs of values from d1 and d2 respectively, and

          missing values represented as default.'''

          res = {}

          for k1, v1 in d1.iteritems():

              if k1 in d2:

                  res[k1] = (v1, d2[k1])

              else:

                  res[k1] = (v1, default)

          if d1 is d2:

              return res

          for k2 in d2:

              if k2 not in d1:

                  res[k2] = (default, d2[k2])

          return res

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				# dicthelpers.py - helper routines for Python dicts
				#
				# Copyright 2013 Facebook
				#
				# This software may be used and distributed according to the terms of the
				# GNU General Public License version 2 or any later version.

				def diff(d1, d2, default=None):
				'''Return all key-value pairs that are different between d1 and d2.

				This includes keys that are present in one dict but not the other, and
				keys whose values are different. The return value is a dict with values
				being pairs of values from d1 and d2 respectively, and missing values
				treated as default, so if a value is missing from one dict and the same as
				default in the other, it will not be returned.'''
				res = {}
				if d1 is d2:
				# same dict, so diff is empty
				return res

				for k1, v1 in d1.iteritems():
				v2 = d2.get(k1, default)
				if v1 != v2:
				res[k1] = (v1, v2)

				for k2 in d2:
				if k2 not in d1:
				v2 = d2[k2]
				if v2 != default:
				res[k2] = (default, v2)

				return res

				def join(d1, d2, default=None):
				'''Return all key-value pairs from both d1 and d2.

				This is akin to an outer join in relational algebra. The return value is a
				dict with values being pairs of values from d1 and d2 respectively, and
				missing values represented as default.'''
				res = {}

				for k1, v1 in d1.iteritems():
				if k1 in d2:
				res[k1] = (v1, d2[k1])
				else:
				res[k1] = (v1, default)

				if d1 is d2:
				return res

				for k2 in d2:
				if k2 not in d1:
				res[k2] = (default, d2[k2])

				return res