3 ms·
Even as simple as fac 1 = 1 fac n = n * (fac (n - 1)) which is a working Haskell implementation? I mean, in Scheme it is longer to write. I enjoy Lisps a
by nine_k 6mo ago
Even as simple as
fac 1 = 1
fac n = n * (fac (n - 1))
which is a working Haskell implementation?
I mean, in Scheme it is longer to write. I enjoy Lisps and use Emacs for everything, but Haskell can be as terse, or even more terse. (Which is not always a good thing.)
- bitwize 6mo agoI think in terms of token count it comes out to about the same; and Lisp admits fewer kinds of tokens.