[LAST ERLANG CLASS] Easier brancy?

[DavideSampietro]

What if we simply impose an order in the reception of the messages from the subtrees? They'll just stack in the mailbox, no deadlock should arise:

brancy(L, R, Fun) ->
	receive
		{ask, P} ->
			% they ask their sons for the values
			L ! {ask, self()},
			R ! {ask, self()},
			% at this point you expect an answer...
			receive
				{L, V1} -> Op1 = V1
			end,
			receive
				{R, V2} -> Op2 = V2
			end,
			% now you evaluate your function for the father
			P ! {self(), Fun(Op1, Op2)}
	end.

This sounds easier to me...

[leonardoarcari]

I thought the same while re-doing the exercise on my own. The only thing I didn't like was having a hanging receive on a branch while the other one may have already finished. The solution I came up with is a little more verbose, but exploiting pattern matching it becomes easier to read and reason about, while keeping the branchy interface the same.

waitBranches(Left, Right, nil, nil) ->
    receive
        {Left, V} -> waitBranches(Left, Right, V, nil);
        {Right, V} -> waitBranches(Left, Right, nil, V)
    end;

waitBranches(Left, Right, LVal, RVal) ->
    receive
        {Left, V} -> {V, RVal};
        {Right, V} -> {LVal, V}
    end.

branchy(F, Left, Right) ->
    receive
        {ask, P} ->
            Left ! {ask, self()},
            Right ! {ask, self()},
            {LVal, RVal} = waitBranches(Left, Right, nil, nil),
            P ! {self(), F(LVal, RVal)}
    end.

Edit: Changed code according to https://github.com/riccardotommasini/ppl/issues/10#issuecomment-356995756

[riccardotommasini]

I like it a lot more, can I include this code in the repository?

[DavideSampietro]

@leonardoarcari you need to swap :) (if I'm getting that right...)

waitBranches(Left, Right, LVal, RVal) ->
    receive
        {Left, V} -> {V, RVal};
        {Right, V} -> {LVal, V}    <-------
    end.

[DavideSampietro]

As for me, no problem with using my code :)

[leonardoarcari]

@DavideSampietro Sure thing! Thanks!