Closed Sintrastes closed 1 year ago
Currently, examples such as foldr (\x y -> x + y) 0 [1,2,3,4,5] do not evaluate properly.
foldr (\x y -> x + y) 0 [1,2,3,4,5]
I believe this has to do with lambdas always having type Dynamic -> Dynamic.
Dynamic -> Dynamic
Currently, examples such as
foldr (\x y -> x + y) 0 [1,2,3,4,5]
do not evaluate properly.I believe this has to do with lambdas always having type
Dynamic -> Dynamic
.