Erlang list matching

erlang

Solution

I think it might not be an error.

The semantics of

[_One, _Two, _Three | _Tail]

is a list of three elements or more.

The semantics of

[_One, _Two | _Tail]

is a list of two elements or more.

Since the third pattern `[ _One, _Two ]` already indicates the case for "a list of two elements", using `[_One, _Two | _Tail]` would be a little bit redundant.

There is a reason for "everything's working as expected". If we place the fourth pattern before the third one, which gives:

many(X) ->
  case X of
    [] ->
      none;
    [_One] ->
      one;
    [_One, _Two | _Tail] ->  %% Switched
      many;
    [_One, _Two] ->          %% Switched
      two
  end.

Then everything won't work as expected. `Mod:many([a,b])` would yield `many` instead of the expected `two`. This is because when the "case" expression is evaluated, X is matched in turn against all the patterns. And this sequential order is guaranteed. The `many` is returned because `[a,b]` matches `[_One, _Two | _Tail]` first, with `_Tail` being `[]`(an empty list).

So, even though `[ _One, _Two | _Tail ]` would work in your case, using `[ _One, _Two , _Three | _Tail ]` is considered a good practice in case you would switch the patterns afterwards.

Problem

I'm working through the book Thinking in Erlang. In "Figure 10: Example of case" it has the following example: ``` many(X) -> case X of [] -> none; [ _One ] -> one; [ _One, _Two ] -> two; [ _One, _Two , _Three | _Tail ] -> many end. ``` It says : If you are wondering why line 9 is not a match against [ _One, _Two | _Tail ], review the list matching rules for list tails at the end of the previous section. But if I actually match against [ _One, _Two | _Tail ] everything still works as expected. Is there an error in the book or am I getting something wrong ?

Original source