I didn't realize it worked until someone mentioned it on the mailing list. (1=)#1 2 3 1 should return 1 1. This form (function;list) of # is basically {y@&x'y}.
Nice find. It was fairly straightforward to implement, but required some tweaks to the parser. Please reopen this issue if it breaks for any reasonable cases.
I didn't realize it worked until someone mentioned it on the mailing list.
(1=)#1 2 3 1
should return1 1
. This form (function;list
) of#
is basically{y@&x'y}
.