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Implicit conversions from Foo? to Foo are a major footgun, because the point of returning a Foo? is to have the caller check for failure before assuming there's
by tlb 2y ago
Implicit conversions from Foo? to Foo are a major footgun, because the point of returning a Foo? is to have the caller check for failure before assuming there's a Foo to use. Even C++ avoids the implicit conversion from optional<T> to T.