3 ms·
> So f[;x][y] is f[x;y]. f[y;x]
by bakul 4y ago
> So f[;x][y] is f[x;y].
f[y;x]
- avmich 4y agoYes. And J has & to bind left or right argument to a dyad making it a monad - e.g. -&5 allows (-&5) 11 which evaluates to 6, while (5&-) 11 evaluates to _6 .