Is scanl1 really partial?

Question

According to the Haskell wiki, the scanl1 function is partial. I don't understand what inputs result in bottom, though. For list functions, I'm used to the problem inputs either being empty lists (like for head) or infinite ones (like for reverse). However, scanl1 seems to handle both of these types of lists correctly. Is this actually a partial function? If so, what's an example of an input that will result in bottom?

Intervention

From the source, it doesn't look partial. Everything seems to be correctly handled. – Warpath

Answer 1

That's a mistake on the wiki. scanl1 is in fact total. (I've updated the wiki.)

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