Compute in one pass 3 tokens position

[Era Scarecrow]

On Tuesday, 1 May 2012 at 02:49:19 UTC, bioinfornatics wrote:
> Le lundi 30 avril 2012 à 14:52 +0200, bioinfornatics a écrit :
> Anyone know how to know the position of 3 token in one sequence 
> in one
> pass?
>
> tok1 = a
> tok2 = b
> tok3 = c
> seq =  blah count me
> b=> 0 a=>2 c=>5
> iterate over sequence if token not yet seen count the number of
> iteration for know in which column it is located. Example own 
> written on
> the fly)

  I would say you are trying to way overcomplicated the solution. 
For simplicity I would use an AA, then a foreach and it will do 
it in 1 pass (2 if you count the inner ones). I'm not sure if 
there's a better solution already in phobos, so I can't refer to 
that :( This could be modified to be more generic so..

//returns AA of tokens that were found and the offset of their 
first occurrence.
int[char] findTokens(const char[] input, char[] tok ...) {
   int[char] offs;
   foreach(i, ch; input) {
     foreach(t; tok) {
       if (ch == t && t !in offs)
         offs[t] = i;
     }
   }
   return offs;
}
unittest {
   assert(findTokens("blah count me", 'a', 'b', 'c')  == ['a':2, 
'b' : 0, 'c':5]);
}