How to expand parameter pack for dependent types?

Question

What is the syntax for expanding a parameter pack where the types I want are dependent on the types directly in the pack?

For example:

template <typename T>
struct foo
{
   typedef T value_type;
};

template <typename ... T>
struct aggregate_of_foo
{
   typedef std::tuple<foo<T>::value_type...> tuple; // MSVC compiler error here
};

Unrelated: can't be just `using tuples = std::tuple;`?? Do not understand why you need the `value_type` of `foo`: — JeJo, Aug 13 '20 at 21:23
@JeJo In this contrived example, yes. but my actual code is more complicated and doesn't follow the pattern that `foo::value_type` will be the same as `T`. — jorgander, Aug 13 '20 at 21:28

score 3 · Answer 1 · answered Aug 13 '20 at 21:11

If you meant foo instead of bar in your typedef, you are just missing the typename keyword before:

template <typename ... T>
struct aggregate_of_foo
{
   using tuple_foo =  std::tuple<typename foo<T>::value_type...>;
                      //         ^^^^^^^^
};

Also note that I named it as tuple_foo because which makes more sense than having just tuple.

score 1 · Accepted Answer · edited Aug 13 '20 at 21:32

1

You have to use the keyword typename because value_type is a dependent name in this context.

template <typename ... T>
struct aggregate_of_foo
{
    typedef std::tuple<typename foo<T>::value_type...> tuple; //here
};

edited Aug 13 '20 at 21:32

cigien

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answered Aug 13 '20 at 21:07

asmmo

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How to expand parameter pack for dependent types?

2 Answers2