Add generic dictionary merge routine.

- Poached from YamlUtils.
- YamlUtils should be updated to use this as the base merge routine.

packages/base/all/vendor-config-onl/src/python/onl/util/__init__.py

@@ -18,3 +18,74 @@ class OnlServiceMixin(object):
         self.logger.critical(msg)
         raise klass(msg)
 
+
+def dmerge(d1, d2):
+    """
+    dictionary merge.
+
+    d1 is the default source. Leaf values from d2 will override.
+
+    d1 is the 'default' source; leaf values from d2 will override.
+    Returns the merged tree.
+
+    Set a leaf in d2 to None to create a tombstone (discard any key
+    from d1).
+
+    if a (sub) key in d1, d2 differ in type (dict vs. non-dict) then
+    the merge will proceed with the non-dict promoted to a dict using
+    the default-key schema ('='). Consumers of this function should be
+    prepared to handle such keys.
+    """
+    merged = {}
+    q = [(d1, d2, merged)]
+    while True:
+        if not q: break
+        c1, c2, c3 = q.pop(0)
+        # add in non-overlapping keys
+        # 'None' keys from p2 are tombstones
+        s1 = set(c1.keys())
+        s2 = set(c2.keys())
+
+        for k in s1.difference(s2):
+            v = c1[k]
+            if type(v) == dict:
+                c3.setdefault(k, {})
+                q.append((v, {}, c3[k],))
+            else:
+                c3.setdefault(k, v)
+
+        for k in s2.difference(s1):
+            v = c2[k]
+            if v is None: continue
+            if type(v) == dict:
+                c3.setdefault(k, {})
+                q.append(({}, v, c3[k],))
+            else:
+                c3.setdefault(k, v)
+
+        # handle overlapping keys
+        for k in s1.intersection(s2):
+            v1 = c1[k]
+            v2 = c2[k]
+
+            if v2 is None: continue
+
+            # two dicts, key-by-key reconciliation required
+            if type(v1) == dict and type(v2) == dict:
+                c3.setdefault(k, {})
+                q.append((v1, v2, c3[k],))
+                continue
+
+            # two non-dicts, p2 wins
+            if type(v1) != dict and type(v2) != dict:
+                c3[k] = v2
+                continue
+
+            if type(v1) != dict:
+                v1 = { '=' : v1, }
+            if type(v2) != dict:
+                v2 = { '=' : v2, }
+            c3.setdefault(k, {})
+            q.append((v1, v2, c3[k],))
+
+    return merged