Re: "is" expression and type tuples

[bearophile via Digitalmars-d-learn]

Jack Applegame:

or use recursion (with splitting in two, and not 1 + n-1).

Bye,
bearophile

I already have one:

template Is(ARGS...) if(ARGS.length % 2 == 0) {
   enum N = ARGS.length/2;
   static if(N == 1) enum Is = is(ARGS[0] : ARGS[1]);

else enum Is = is(ARGS[0] : ARGS[N]) && Is!(ARGS[1..N], ARGS[N+1..$]);

}

That's 1 + n-1 :-)

Bye,
bearophile