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return self.rev_id
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def _any_real_changes(self):
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"""Are there real changes between new_inventory and basis?
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For trees without rich roots, inv.root.revision changes every commit.
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But if that is the only change, we want to treat it as though there
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new_entries = self.builder.new_inventory.iter_entries()
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basis_entries = self.basis_inv.iter_entries()
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new_path, new_root_ie = new_entries.next()
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basis_path, basis_root_ie = basis_entries.next()
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# This is a copy of InventoryEntry.__eq__ only leaving out .revision
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def ie_equal_no_revision(this, other):
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return ((this.file_id == other.file_id)
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and (this.name == other.name)
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and (this.symlink_target == other.symlink_target)
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and (this.text_sha1 == other.text_sha1)
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and (this.text_size == other.text_size)
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and (this.text_id == other.text_id)
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and (this.parent_id == other.parent_id)
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and (this.kind == other.kind)
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and (this.executable == other.executable)
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if not ie_equal_no_revision(new_root_ie, basis_root_ie):
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for new_ie, basis_ie in zip(new_entries, basis_entries):
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if new_ie != basis_ie:
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# No actual changes present
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def _check_pointless(self):
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if self.allow_pointless:
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# A merge with no effect on files
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if len(self.parents) > 1:
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# work around the fact that a newly-initted tree does differ from its
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# Shortcut, if the number of entries changes, then we obviously have
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if len(self.builder.new_inventory) != len(self.basis_inv):
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if (len(self.builder.new_inventory) != 1 and
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self.builder.new_inventory != self.basis_inv):
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# If length == 1, then we only have the root entry. Which means
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# that there is no real difference (only the root could be different)
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if (len(self.builder.new_inventory) != 1 and self._any_real_changes()):
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raise PointlessCommit()