In template void f(){ typename T::value_type v{}; }, why is typename required?

Because T::value_type is a dependent qualified name that could be a type or a value

Because value_type is a private member that cannot be named without the disambiguator

Because T has no default constructor, so its members must be named as types explicitly

Because typename enables SFINAE on the dependent member when substitution occurs

Multiple ChoiceHard 45s

typename dependent qualified name

In template<class T> void f(){ typename T::value_type v{}; }, why is typename required?

Options

1Because T::value_type is a dependent qualified name that could be a type or a value
2Because value_type is a private member that cannot be named without the disambiguator
3Because T has no default constructor, so its members must be named as types explicitly
4Because typename enables SFINAE on the dependent member when substitution occurs

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