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mutually recursive functions and tail recursion
f#

I have the following function:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

 

        let rec build_vars notFlag (tokens: string list) (filter_accessor: IFilterManager) (vars:(IfLinkType option)*(bool*FilterExpression) list) =
            match tokens with
            | "not"::var::tail -> build_vars true (var::tail) filter_accessor vars
            | var::[] -> 
                match fst vars with
                | None -> IfLinkType.Or, [(notFlag, new FilterExpression(var, filter_accessor))]
                | Some any -> any, snd vars @ [(notFlag, new FilterExpression(var, filter_accessor))]
            | var::"and"::var2::tail -> 
                match fst vars with
                | Some any when any <> IfLinkType.And -> raise (TemplateSyntaxError ("'if' tags can't mix 'and' and 'or'"))
                | _ -> ()
                build_vars false (var2::tail) filter_accessor (Some IfLinkType.And, snd vars @ [(notFlag, new FilterExpression(var, filter_accessor))])
            | var::"or"::var2::tail -> 
                match fst vars with
                | Some any when any <> IfLinkType.Or -> raise (TemplateSyntaxError ("'if' tags can't mix 'and' and 'or'"))
                | _ -> ()
                build_vars false (var2::tail) filter_accessor (Some IfLinkType.Or, snd vars @ [(notFlag, new FilterExpression(var, filter_accessor))])
            | _ -> raise (TemplateSyntaxError ("invalid conditional expression in 'if' tag"))

options 3 and 4 from this match beg to be separated out into another function. Which brings two questions:

1. what would be the syntax for these two functions recursively calling each other?

2. will it cease to be the tail recursion? I suspect that this is the case, so probably it is better left the way it is now even though it is more verbose

.

See

[link:stackoverflow.com]

for the syntax. F# does tail calls (more general than tail recursion), so for example the F and G functions in the answer linked above comprise a 'loop' (no stack overflow) since each calls the other in tail position.

By on 5/11/2009 9:49 AM ()

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