> Can you post a complete, runnable example that illustrates your problem?

Strange as it is, now it works here too... - I don't know, what went wrong yesterday. Thanks anyway. :-)

On Friday, 14 September 2018 at 20:43:45 UTC, SrMordred wrote:
> On Friday, 14 September 2018 at 19:44:37 UTC, berni wrote:
>> a) I've got an int[] which contains only 0 und 1. And I want to end with a string, containing 0 and 1. So [1,1,0,1,0,1] should become "110101". Of course I can do this with a loop and ~. But I think it should be doable with functional style, which is something I would like to understand better. Can anyone help me here? (I think a.map!(a=>to!char(a+'0')) does the trick, but is this good style or is there a better way?)

Yes, that's fine. That gives you a range; if you want an array, call array() with that expression. Also, since you know the addition won't overflow, you can substitute the to!char call with a cast (arr.map!(c => cast(char)(c + '0')).array).

Another option is to index a string literal: arr.map!(c => "01"[c]).array

> auto a = [1,0,1,1,1,0,1,0,1,1,1,1,0];
>     a.map!(to!string)
>      .join("")
>      .chunks(4)
>      .map!(to!string) //don´t know why the chunks are not already strings at this point ;/
>      .writeln;

That's needlessly complicated and inefficient; this will allocate one string per array element.

If you don't need the result to be actual arrays, this works:

	auto res = [1,0,1,1,1,0,1,0,1,1,1,1,0]
		.map!(c => "01"[c])
		.chunks(4);
	assert(equal!equal(res, ["1011","1010","1111","0"]));

If you need a real array of arrays:

	auto arr = res
		.map!array
		.array;
	assert(arr == ["1011","1010","1111","0"]);