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And this is meaningful only in the context of a lazy language. In the context of strict, statically typed language an algebraic Option type is good-enough.
by throwaway487549 8y ago
And this is meaningful only in the context of a lazy language.
In the context of strict, statically typed language an algebraic Option type is good-enough.
- gizmo686 8y agoMaybe is an algebraic Option type. It just happens to implement an interface that abstracts away the if(None) checks that you would be doing anyway. Lazyness has nothing to do with it.