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Hi
suma [] = 0 suma (h:t) = h + suma t suma [] = 0 suma h:t = h + suma t
suma [] = 0 suma (h:t) = h + suma t
suma [] = 0 suma h:t = h + suma t
Infix operators bind less tightly than function application, so the compiler sees: (suma h) : t = h + (suma t) Hence the compiler gets confused. Thanks Neil
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