On Saturday 05 September 2015 22:22, Namal wrote:
> import std.stdio, std.algorithm, std.array;
>
> bool abundant(int n){
[...]
> }
>
>
>
> void main(){
>
> long sum;
> int[] arr;
> int[28123] mark;
>
> foreach(i;1..28124)
> if(abundant(i))
> arr~=i;
[...]
> }
>
> How can I generate the array arr the functional way with reduce
> and the function abundant as filter. Tried it on my own but
> failed.
import std.range: iota;
int[] arr = iota(1, 28124).filter!abundant.array;
> Also, if I use reduce on an array, like I did for the
> sum, what does the zero mean in the range?
You mean the 0 in this: `reduce!((a,b)=>a+b)(0,a)`, right? It's used as the
initial value for the `a` argument to the lambda. That is, the first
calculation there is 0+a[0]. Then a[1] is added to that, and so on.
Note that there's a specialized `std.algorithm.iteration.sum`.
