Question

Hi I’m learning Erlang via Learn You Some Erlang by Fred Hebert.

And I’ve come across a code that I’m confuse about:

sword(1) -> throw(slice);
sword(2) -> erlang:error(cut_arm);
sword(3) -> exit(cut_leg);
sword(4) -> throw(punch);
sword(5) -> exit(cross_bridge).

talk() -> "blah blah".

black_knight(Attack) when is_function(Attack, 0) ->
    try Attack() of
        _ -> "None shall pass."
    catch
        throw:slice -> "It is but a scratch.";
        error:cut_arm -> "I've had worse.";
        exit:cut_leg -> "Come on you pansy!";
        _:_ -> "Just a flesh wound."
    end.

So here’s the confusion. I don’t understand sword(#) function. Why are there number as parameter? The function is_function actually check if these function are of arity 0 and apparently all the sword(#) functions are of arity 0.

Also the way to pass in the sword(#) function to the black_knight function is different compare to the talk function.

Here’s how the book pass a sword function and the talk function.

exceptions:black_knight(fun exceptions:talk/0).

vs

exceptions:black_knight(fun() -> exceptions:sword(1) end).

The talk function we just pass the function where as the sword(1) function we have to wrap it with a anonymous function. I don’t get it.

So the questions are:

1. Why is passing these sword(#) different from talk function.
2. Why sword(#) have a number as a parameter?
3. Why sword(#) have 0 arity when it seems like it have an arity of 1 (I’m counting the number parameter as a parameter)?

The chapter of the book I’m at.

Thank you for your time.

Answer 1

1. If you look at the guard statement for the black_knight function, is_function(Attack, 0), it will only match the definition if the function passed in takes 0 parameters. Since talk takes 0 parameters, it can be passed in directly. sword takes one parameter, so you need to wrap it in an anonymous function that takes 0 parameters before you can pass it in.
2. The number in the definition of each clause is an example of pattern matching. If you call sword with 1 as the argument, you will execute the code in the clause sword(1) ->. If you pass in 2 as the argument, you will execute the clause sword(2) ->. See this section in Learn You Some Erlang for a more complete description.
3. sword does have an arity of 1, so you were counting parameters correctly.