sort values of an orddict - sorting

Question

In order to extract the values (records) of an orddict as a sorted list, tried this:
-module(test).
-compile(export_all).
-record(node, {name="", cost=0}).
test() ->
List = orddict:append("A",#node{name="A",cost=1},
orddict:append("B",#node{name="B",cost=2},
orddict:new())),
lists:sort(fun({_,A},{_,B}) -> A#node.cost =< B#node.cost end,
orddict:to_list(List)).
The sort fails with exception error: {badrecord,node}.
What would be the correct syntax?
Solved:
The correct insertion method is orddict:store/2 instead of orddict:append/2. Then the pattern {_,A} matches for the comparison function.

Answer 1

The correct syntax is:
lists:sort(fun({_,[A]},{_,[B]}) -> A#node.cost =< B#node.cost end,
orddict:to_list(List)).
I not found note about this in documentation,but you can look in source code of module.
As #Pascal write in comments the reason is that orddict:append/3 is a function provided to append a value to an existing Key/Value pair where Value must be a list. In the use case, the key doesn't exist, so the pair is created and the Value append to an empty list.
Btw, you always can print and compare real and expected result.
io:format("~p~n",[orddict:to_list(List)])
For your example that is:
[{"A",[{node,"A",1}]},{"B",[{node,"B",2}]}]