10 Sep
2013
10 Sep
'13
6:44 p.m.
Hm, maybe:
bool :: a -> a -> Bool -> a
cond :: (a -> b) -> (a -> b) -> (a -> Bool) -> a -> b
On 11 September 2013 00:36, Christopher Done
+1
I use:
bool :: (a -> b) -> (a -> b) -> (a -> Bool) -> a -> b
which is better for point-free programming, but I'll go along with yours because bool :: a -> a -> Bool -> a is also useful.