Op vrijdag 24 oktober 2014 15:32:33 UTC+2 schreef Laurens Van Houtven:
>
> Hi Roelof, 
>
> On 24 Oct 2014, at 15:12, Roelof Wobben <[email protected] <javascript:>> 
> wrote: 
>
> > I understand that part  but when I look at the map part there is no x. 
> When I look at the function no x. 
>
> I’m assuming you mean that when you look at the function, you do see the x 
> — it’s right there, both in the argument list and in the body. 
>
> > So how does Clojure know what the value of x is. 
>
> Then you haven’t quite understood it ;-) 
>
> Consider: 
>
> (defn double [x] (* x 2)) 
>
> i.e. a function that doubles its argument. When you call (double 4), how 
> does Clojure “know” what x is? The same way that you call double with 4, 
> map applies the function you pass it to each element of the collection you 
> pass it. So, for example: (map double [3 5 7]) will basically do [(double 
> 3) (double 5) (double 7)]. The “x” is “replaced”, if you will, just like it 
> is with regular function application. That’s all map is: a bunch of 
> function applications and tying the results together :-) 
>
> *(except that it actually returns a lazy seq, not a vector, but that’s 
> besides the point here). 
>
> hth 
> lvh 
>


I understand that part. 

so we have (map second-item collection)  where second-item and collection 
are arguments of map. 
Then we have second-item ( fn [x] (get x 2)    
which can be read as : 

( second-item [x] (get x 2)  .

in the map there is only ( map second-item collection) 

or is the collection a argument of second-item and a argument of map. 

Roelof

   

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