more robust assignment of lineno for keyword args

jeremyhylton · jeremyhylton · commit 3d9f5e4de2f3 · 2000-02-16T00:51:37.000Z
get the lineno from the name of the keyword arg

example of case that didn't work--
def foo(x, y, a = None,
	b = None):
diff --git a/Lib/compiler/transformer.py b/Lib/compiler/transformer.py
@@ -993,9 +993,9 @@ def com_argument(self, nodelist, kw):
       n = n[1]
     if n[0] != token.NAME:
       raise SyntaxError, "keyword can't be an expression (%s)"%n[0]
-    n = Node('keyword', n[1], result)
-    n.lineno = result.lineno
-    return 1, n
+    node = Node('keyword', n[1], result)
+    node.lineno = n[2]
+    return 1, node
 
   def com_subscriptlist(self, primary, nodelist, assigning):
     # slicing:      simple_slicing | extended_slicing
diff --git a/Tools/compiler/compiler/transformer.py b/Tools/compiler/compiler/transformer.py
@@ -993,9 +993,9 @@ def com_argument(self, nodelist, kw):
       n = n[1]
     if n[0] != token.NAME:
       raise SyntaxError, "keyword can't be an expression (%s)"%n[0]
-    n = Node('keyword', n[1], result)
-    n.lineno = result.lineno
-    return 1, n
+    node = Node('keyword', n[1], result)
+    node.lineno = n[2]
+    return 1, node
 
   def com_subscriptlist(self, primary, nodelist, assigning):
     # slicing:      simple_slicing | extended_slicing
