这个Haskell函数如何工作？

Question

I was struggling to see how this function worked. For the nth number it should calculate the sum of the previous three elements.

f' :: Integer->Integer

f' = helper 0 0 1

 where

   helper a b c 0 = a

   helper a b c n = helper b c (a+b+c) (n-1)

Thanks for your time

Answer 1

Perhaps the part that you're missing is that

f' = helper 0 0 1

is the same thing as

f' x = helper 0 0 1 x

Otherwise, see Dave's answer.

Answer 2

It's a fairly simple recursive function. When called with three elements (I'm guessing seeds for the sequence) and a number of terms, it calls itself, cycling the seed left by one and adding the new term (a+b+c). When the "number of steps remaining" counter reaches 0, the edge case kicks in and just returns the current sequence value. This value is passed back up all the function calls, giving the final output.

The f' function provides a simple wrapper around the helper function (which does the work I described above), providing a standard seed and passing the requested term as the 4th parameter (MathematicalOrchid explains this nicely).

Answer 3

say its called with f' 5

below is the sequence in which it will get executed:

iteration 1: helper 0 0 1 5

iteration 2: helper 0 1 (0+0+1) 4

iteration 3: helper 1 1 (0+1+1) 3

iteration 4: helper 1 2 (1+1+2) 2

iteration 5: helper 2 4 (1+2+4) 1

iteration 6: helper 4 7 (2+4+7) 0 => 4

Answer 4

It is like a Fibonacci sequence, but for 3 numbers, not 2:

F'_n = F'_{n-1} + F'_{n-2} + F'_{n-3}

where Fibonacci sequence is

F_n = F_{n-1} + F_{n-2}