How can I extend typed Arrays in Swift? -

How can I extend typed Arrays in Swift? -

how can extend swift's array<t> or t[] type custom functional utils?

browsing around swift's api docs shows array methods extension of t[], e.g:

extension t[] : arraytype { //... init() var count: int { } var capacity: int { } var isempty: bool { } func copy() -> t[] }

when copying , pasting same source , trying variations like:

extension t[] : arraytype { func foo(){} } extension t[] { func foo(){} }

it fails build error:

nominal type t[] can't extended

using total type definition fails use of undefined type 't', i.e:

extension array<t> { func foo(){} }

and fails array<t : any> , array<string>.

curiously swift lets me extend untyped array with:

extension array { func each(fn: (any) -> ()) { in self { fn(i) } } }

which lets me phone call with:

[1,2,3].each(println)

but can't create proper generic type extension type seems lost when flows through method, e.g trying replace swift's built-in filter with:

extension array { func find<t>(fn: (t) -> bool) -> t[] { var = t[]() x in self { allow t = x t if fn(t) { += t } } homecoming } }

but compiler treats untyped still allows calling extension with:

["a","b","c"].find { $0 > "a" }

and when stepped-thru debugger indicates type swift.string it's build error seek access string without casting string first, i.e:

["a","b","c"].find { ($0 string).compare("a") > 0 }

does know what's proper way create typed extension method acts built-in extensions?

after while trying different things solution seems remove <t> signature like:

extension array { func find(fn: (t) -> bool) -> [t] { var = [t]() x in self { allow t = x t; if fn(t) { += t } } homecoming } }

which works intended without build errors:

["a","b","c"].find { $0.compare("a") > 0 }

arrays swift