Type inference for recursive functions can be challenging:
factorial(integer n) = if n then n * factorial(n-1) else 1
When factorial(n-1) is called, the type may only be no_type at that point. Then, the result of using a no_type operand must be promoted to the other operand's type. So, integer_type * no_type results in integer_type.
rbbarbosa
commented
5 hours ago
Type inference for recursive functions can be challenging:
factorial(integer n) = if n then n * factorial(n-1) else 1When
factorial(n-1)is called, the type may only beno_typeat that point. Then, the result of using ano_typeoperand must be promoted to the other operand's type. So,integer_type * no_typeresults ininteger_type.