5 ms·
In there is also ApplicativeDo that works nicely with this. do x <- f1 y <- f2 return $ x + y this is evaluated as applicative in same
by raluk 1y ago
In there is also ApplicativeDo that works nicely with this.
do
x <- f1
y <- f2
return $ x + y
this is evaluated as applicative in same way.