Erlang (programming language)/Tutorials/Folding

Fun with folding

Fold is a powerful tool, when you get to know it. lists:foldr(F,S,L) takes three arguments: F is some folding function, S is some starting value, and L is some List that needs folding. Let us do some simple folding. The following fold calculates the lenght of a list. Here the current list item, _C is ignored, and the accumulator, A, counts the number of elements in the list.

 1> lists:foldr( fun(_C,A)->A+1 end, 0, [1,2,3]).
 3

We could reverse a list with fold if we like.

 2> lists:foldr(fun(C,A)->A++[C] end, [], [a,b,c]). 
 [c,b,a]

Or to get fancy we could try a finite difference.

 3> lists:foldr(
   fun(C,{P,A})->{C,[P-C]++A} end, 
   {0,[]},   
   [1,4,9,16]).  

 {1,[3,5,7,-16]}

In example 3, we used a tuple to remember the previous value so we could use it for next difference. Fold is a function with a limited attention span. We might like to delete the last item because is has a non-intuitive relation to the source list.

Calculating a continuted fraction with fold seems natural

lists:foldr(fun(C,A)->(A+1)/C end, 1, lists:seq(1,1000)).  
1.718281828459045

The answer appears to be the value e-1, which we can verify with:

math:exp(1) - 1. 
1.718281828459045