Erlang list matching

Question

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 ?

Ning Ning · Accepted Answer · 2012-06-04T22:00:56

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.

Erlang list matching

2 Answers