CODEWITHSUNDEEP
listerlangtuples
Jazzbaron
Jazzbaron

Reputation: 75

[ERLANG]: Why answer returns list in a list rather than list?

Given the module 

-module(p1). 

-export([f2/2]). 

f2([A, A | B]) -> {A, B}; 
f2([A, B | _]) -> {A, B}; 
f2([A]) -> A; 
f2(_) -> no_match.

I was asked to try the input value 

 p1:f2([1,1,[1,1]]).

My answer would be that this input value matches the first function clause of f2, and as such gives us the result:

{1,[1,1]}

But according to the given answer sheet, the answer is

{1,[[1,1]]}

I can't quite get my head around why we get a list within a list, rather than a list in the answer. Would appreciate an explanation for this, thank you.

Upvotes: 0

Views: 74

Answers (1)

mpm
mpm

Reputation: 3584

You should look into how head-tail works in Erlang. Tail is always a list; could be a one element list, could be an empty list, but always a list.

Lets look into example in shell:

2> Fun = fun([H | T]) -> {H, T} end. 
#Fun<erl_eval.6.90072148>
3> Fun([a]).
{a,[]}
4> Fun([a,b]).
{a,[b]}
5> Fun([a,[a,b]]).
{a,[[a,b]]}
6> Fun([a,b,c]).  
{a,[b,c]}
7> Fun([a,b,c,d]).
{a,[b,c,d]}

So in you case what you are returning is A which is repeted 1 and B which is list of all remaining elements. In our case only one element remains, so we return one-element list. And this element is a two-element list, hence list in a list [[1, 1]].

Upvotes: 2

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