Re: "is" expression and type tuples

[Jack Applegame via Digitalmars-d-learn]

On Tuesday, 3 March 2015 at 16:42:22 UTC, bearophile wrote:

But it should be not too much hard to implement it your code. Just use two is(), 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..$]);

}