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CHAPTER 8 MA STERING F#: C OMM ON TE CHNIQUE S
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Here is a correct implementation: let rec mapAcc f inputList acc = match inputList with | [] -> List.rev acc | h::t -> mapAcc f t (f h :: acc) let map f inputList = mapAcc f inputList [] > map (fun x -> x * x) [1;2;3;4];; val it : int list = [ 1; 4; 9; 16]
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Tail Recursion and Object-Oriented Programming
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You will often need to implement object members with a tail-recursive implementation. For example, consider the following list-like data structure: type Chain = | ChainNode of int * string * Chain | ChainEnd of string member chain.LengthNotTailRecursive = match chain with | ChainNode(_,_,subChain) -> 1 + subChain.LengthNotTailRecursive | ChainEnd _ -> 0 The implementation of the LengthNotTailRecursive is not tail recursive, simply because the addition 1 + applies to the result of the recursive property invocation. One obvious tailrecursive implementation uses a local recursive function with an accumulating parameter, as shown in Listing 8-9. Listing 8-9. Making an Object Member Tail Recursive type Chain = | ChainNode of int * string * Chain | ChainEnd of string // The implementation of this property is tail recursive. member chain.Length = let rec loop c acc = match c with | ChainNode(_,_,subChain) -> loop subChain (acc+1) | ChainEnd _ -> acc loop chain
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C HAPTE R 8 MAS TE RING F#: COMMON TEC HN IQUES
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Note The list-processing functions in the F# library module Microsoft.FSharp.Collections.List
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are tail recursive, except where noted in the documentation, and some of them have implementations that are specially optimized to take advantage of the implementation of the list data structure.
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Tail Recursion and Processing Unbalanced Trees
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In the section, we consider tail-recursion problems that are much less common in practice but where it is important to know what techniques to apply should you require them. The techniques also illustrate some important aspects of functional programming, in particular an advanced technique called continuation passing. Tree-structured data is generally more difficult to process in a tail-recursive way than liststructured data. For example, consider the following tree structure: type Tree = | Node of string * Tree * Tree | Tip of string let rec sizeNotTailRecursive tree = match tree with | Tip _ -> 1 | Node(_,treeLeft,treeRight) -> sizeNotTailRecursive treeLeft + sizeNotTailRecursive treeRight The implementation of this function is not tail recursive. Luckily, this is rarely a problem, especially if you can assume that the trees are balanced. A tree is balanced when the depth of each subtree is roughly the same. In that case, a tree of depth 1,000 will have about 21000 entries. Even for a balanced tree of this size, the recursive calls to compute the overall size of the tree will not recurse to a depth greater than 1,000 not deep enough to cause stack overflow except when the routine is being called by some other function already consuming inordinate amounts of stack. Many data structures based on trees are balanced by design; for example, the Set and Map data structures implemented in the F# library are based on balanced binary trees. However, some trees can be unbalanced; for example, you can explicitly make a highly unbalanced tree: let rec mkBigUnbalancedTree n tree = if n = 0 then tree else Node("node",Tip("tip"),mkBigUnbalancedTree (n-1) tree) let let let let let let tree1 tree2 tree3 tree4 tree5 tree6 = = = = = = Tip("tip") mkBigUnbalancedTree mkBigUnbalancedTree mkBigUnbalancedTree mkBigUnbalancedTree mkBigUnbalancedTree
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tree1 tree2 tree3 tree4 tree5
CHAPTER 8 MA STERING F#: C OMM ON TE CHNIQUE S
Calling tree6.Size now risks a stack overflow. You can solve this in part by trying to predict whether the tree will be unbalanced to the left or right and by using an accumulating parameter: let rec sizeAcc acc tree = match tree with | Tip _ -> 1+acc | Node(_,treeLeft,treeRight) -> let acc = sizeAcc acc treeLeft sizeAcc acc treeRight let size tree = sizeAcc 0 tree This algorithm works for tree6, because it is biased toward accepting trees that are skewed to the right. The recursive call that processes the right branch is a tail call, which the call that processes the left branch is not. This may be OK if you have prior knowledge of the shape of your trees. However, this algorithm still risks a stack overflow, and if necessary, you may have to change techniques. One way to do this is to use a much more general and important technique known as continuation passing.
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