3 ms·
Ah, but Haskell is lazy so loop will never be evaluated and a loop >> constantly_unit will return unit no matter what.
by wz1000 12y ago
Ah, but Haskell is lazy so loop will never be evaluated and a
loop >> constantly_unit
will return unit no matter what.
- tel 12y agoAh, yeah, I was being fast and loose with laziness. You have to `seq` the argument to `constantly_unit`, too.