optimization - Least expensive way to construct cyclic list in Haskell -

so if want construct circular list of n 0's , 1 1, of following ways better/cheaper? , there better/cheaper way? taking account n integer , may large (though realistically not going exceed 2^32).

azerosandones :: integer -> [int] azerosandones n      | n >= 0    = cycle (genericreplicate n 0 ++ [1])     | otherwise = [] 

versus

bzerosandones :: integer -> [int] bzerosandones n      | n >= 0    = tail (cycle (1 : genericreplicate n 0))     | otherwise = [] 

i'd go second, because it's efficient , plenty clear enough. first depend on whether genericreplicate able fuse ++ in fashion. best way find out sure run

ghc -o2 -ddump-simpl -dsuppress-all whatever.hs | less 

and pore on spews.

---

that said, entire length of cycle allocated in memory. such nature of cycle function implemented, , not seem change (barring significant advance—foldr/build fusion not seem sufficient). better off avoiding altogether writing rest of code differently.

---

ah yes, thought of else. if consume list in "single-threaded" fashion, can dispense cycle altogether:

weirdthing n = genericreplicate n 0 ++ [1] ++ weirdthing n

and that's final answer. makes infinite list instead of cyclic list, when n big enough, that's better.