"is" expression and type tuples

[bearophile via Digitalmars-d-learn]

Jack Applegame:

> On Tuesday, 3 March 2015 at 17:49:24 UTC, bearophile wrote:
>> That's 1 + n-1 :-)
> Could you please explain what does '1 + n-1' mean?

This is your code:

>> 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..$]);
>> }

The recursion scheme you are using is working on a single item (a 
single pair of items), and then calling the recursion on all 
other items but the first.

If you look in std.traits you see examples of a different 
recursion that reduces compilation time:

template isExpressionTuple(T ...)
{
     static if (T.length >= 2)
         enum bool isExpressionTuple =
             isExpressionTuple!(T[0 .. $/2]) &&
             isExpressionTuple!(T[$/2 .. $]);
     else static if (T.length == 1)
         enum bool isExpressionTuple =
             !is(T[0]) && __traits(compiles, { auto ex = T[0]; });
     else
         enum bool isExpressionTuple = true; // default
}


Bye,
bearophile