Closed deepak1556 closed 9 years ago
No, compose(f, g)(x)
is f(g(x))
, as expected.
The outer transducer (returned by f) processes the item before passing it to the inner transducer (returned by g). Roughly like this:
g(x): g-ing, then x
f(x): f-ing, then x
f(g(x)): f-ing, then g-ing, then x
Ahh thanks!
compose(f, g) //=> g(f(x)) Is this the intended behaviour or should it be the other way ?