Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms / 3 / c / Solution

The three outcomes, in printed order, are ZagType 'A', ZigType undefined, and a static type error.

In the first expression, pattern matching discovers that the first tail is nonempty, so the ignored first label, the bottom value $\perp$, is never demanded. The singleton Cans 'A' Nil returns ZagType 'A'.

In the second expression the tail is Nal, so the result is ZigType applied to the bottom value. Under lazy evaluation this has a defined outer constructor and an undefined field: case lastZig (Cins undefined Nal) of ZigType _ -> True yields True, whereas extracting or fully printing the payload fails. Thus the result is $\mathrm{ZigType}(\perp )$, not a wholly undefined outer value.

The original PDF really prints Nil in the third expression. Cins requires a tail of type Zag b a, whereas Nil has type Zig c d. These distinct type constructors cannot be unified. There is no well-typed expression to evaluate, regardless of the undefined first label; replacing Nil by Nal would change the question.