Re: [xsl] How do I get a SUM of the string-length of all child nodes ??

[Mukul Gandhi <mukul_gandhi@xxxxxxxxx>]

Thank you Dimitre for explanation ;) I am obliged ..

Regards,
Mukul

--- Dimitre Novatchev <dnovatchev@xxxxxxxxx> wrote:
> >  What does map function do? 
> 
> Formal definition in Haskell prelude (Prelude.hs):
> 
>      map              :: (a -> b) -> [a] -> [b]
>      map f xs          = [ f x | x <- xs ]
> 
> 
> The first line above defines the type (signature) of
> the "map"
> function. It takes two arguments:
> 
>    -  a function of type a -> b (with domain "a" and
> codomain "b"  --
> this means that the
>       type of thie arguments is "a" and the type of
> the results is "b")
>       Because "map" has an argument, which is a
> function, "map" is a
> higher-order
>      function.
> 
>   - a list of elements each of type "a"
> 
> The type of the result is a list of elements of type
> "b".
> 
> The second line of the definition above defines
> exactly the map
> function. We see that the result of applying a
> function "map" to its
> two arguments -- a function "f" and a list "xs" -- 
> is the set of all
> "f x"  (which means f(x))  where "x" belongs to the
> list "xs"
> 
> More informally, we have a list of elements of the
> same type ("a") and
> a function "f", defined on "a" and producing results
> of type "b".
> 
> The result of applying "map" on "f" and a list "xs"
> (all of whose
> elements are of type "a") is another list  "ys" ,
> whose elements "y"
> are the results of applying "f" on the corresponding
> elements "x" of
> "xs".
> 
> 
> Example:
> 
>   map (2 * )  [1,2,3,4]  =  [2,4,6,8]
> 
> where (2 *) is a function, which produces twice its
> argument.
> 
>   map string-length ['one', "two", "three", "four"] 
> =  [3, 3, 5, 4]
> 
> then, we'll have:
> 
>   sum (map string-length ['one', "two", "three",
> "four"] ) = 15
> 
> 
> > Please explain what does
> > sum(f:map(f:string-length(), /*/node())) mean ..
> 
> Almost the same as the last line above  -- I hope it
> is clear now.
> 
> 
> Cheers,
> Dimitre Novatchev.
> 
> 


		
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