Cohen-Sutherland Clipping
Taken from Wikipedia's article on Cohen-Sutherland:


  Both endpoints are in different regions: in case of this nontrivial situation the algorithm finds one of the two points that is outside the viewport region (there will be at least one point outside). The intersection of the outpoint and extended viewport border is then calculated (i.e. with the parametric equation for the line), and this new point replaces the outpoint. The algorithm repeats until a trivial accept or reject occurs.


So yes, since your new first point is inside the viewport, but your second point is outside, the situation is non-trivial (i.e. no "go render it all" or "go discard it all"), it will repeat with the second point. Afterwards, both points will be inside the viewport, therefore it is a trivial draw and no further action is needed.