[swift-evolution] Proposal: Implement == and < for tuples where possible, up to some high arity

Andrew Bennett cacoyi at gmail.com
Mon Dec 7 18:29:56 CST 2015

I'd love to see a type like tuple<(A,B,C),D>, that is equivalent to
Also you could have functions on the tuple:

for (A,B,C,D,E,F):

func prefix<A,B,C,...>() -> (A,B,C)
func suffix<...,D,E,F>() -> (D,E,F)

Then you may be able to do something like this:
func ==<A:tuple<>, B: tuple<>, where A: Equatable, B:Equatable>(lhs:
tuple<A,B>, rhs: tuple<A,B>) {
    return ls.prefix() as A == rhs.prefix() as A && lhs.suffix() as B ==
rhs.suffix() as B
extension tuple<Equatable,Equatable>: Equatable {}

The downside I can see for that is that it will match the same tuple n
different ways. You can overcome this by only allowing tuple<A,B> to append
a single element, but it's less useful. Alternatively it will always match
the function with the most elements in the left-most argument, and cannot
match against the empty tuple.

On Tue, Dec 8, 2015 at 11:18 AM, Dmitri Gribenko via swift-evolution <
swift-evolution at swift.org> wrote:

> On Mon, Dec 7, 2015 at 4:17 PM, Brent Royal-Gordon
> <brent at architechies.com> wrote:
> >>>>      return zip([a1,b1,c1,d1,e1],
> [a2,b2,c2,d2,e2]).lazy.filter(==).first.flatMap(<) ?? false
> >>>>
> >>>> Okay, so that wasn’t quite as easy as I thought when I started writing
> >>>> the email.
> >>>
> >>> That also allocates two intermediate arrays (the inputs to zip()).
> >>
> >> And relies on all variables having the same type.
> >
> > Both true. What actually happened was, when I looked at the tuple
> version, I thought “okay, that’s basically just zipping the two together
> and then running < on each pair". In trying to write the email making that
> point, I discovered that this “easy” alternative is actually ridiculously
> complicated once you translate it from a vague notion to running code, and
> thought that was funny.
> We have lexicographicalCompare(), so it is not that complicated after all.
> Dmitri
> --
> main(i,j){for(i=2;;i++){for(j=2;j<i;j++){if(!(i%j)){j=0;break;}}if
> (j){printf("%d\n",i);}}} /*Dmitri Gribenko <gribozavr at gmail.com>*/
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