Reducing an array

[MattCoder via Digitalmars-d-learn]

On Saturday, 19 April 2014 at 17:12:10 UTC, Ali Çehreli wrote:
> On 04/19/2014 09:55 AM, MattCoder wrote:
>
> > On Friday, 18 April 2014 at 20:02:41 UTC, Tim Holzschuh via
> > Digitalmars-d-learn wrote:
> > void main(){
> >      int myfilter(int a){
> >          static int[] b;
>
> That static variable makes this solution non-reentrant...

Yes, you're completely right and I already knew that. But 
remember Like I said previously, I would like to convert that 
code to something close to "extensions" in C#, in that case I can 
take the address of the array to check if it's a new Filter or 
not. for example, current in D I can do this:

import std.stdio;
import std.array;
import std.range;
import std.algorithm;

void main(){
	int myfilter(int[] *addr, int a){
		static int[] b;
		static int[] *address;
		
		if(address != addr){
			address = addr;
			b.clear();
		}
		
		if(!b.canFind(a)){
			b.insertInPlace(b.length, a);
			return a;
		}
		return -1;
	}
	
	auto arr  = [1,1,2,3,2,2,4,5,5,1];	
	auto arr2 = [3,4,3,9,9,7,4,4,4,7];

	auto arrFiltered = arr.filter!(a => myfilter(&arr, a) >= 0);
	writeln(arrFiltered);
	
	auto arrFiltered2 = arr2.filter!(a => myfilter(&arr2, a) >= 0);
	writeln(arrFiltered2);	
}

But with extensions, the argument "address" (i.e: &arr) on the 
calling of myfilter can be avoided!

Matheus.