Why is it not possible to use a template alias in function as parameter and to be automatically deduced?

Question

I was trying to simplify a template function by using a template alias instead of the underlying type:

template<typename T,
    typename Traits = std::char_traits<T>,
    typename Allocator = std::allocator<T>,
    typename String = std::basic_string<T, Traits, Allocator>,
    typename Vector = std::vector<String>>
Vector make_vector_from_string(const String & str)
{
    //do something with str parameter

    return Vector{
        {str}
    };
}

But the caller is required to specify the template type because the compiler fails to deduce T for the parameter:

std::string bar{"bar"};
auto strings{make_vector_from_string<char>(bar)};

If the function parameter type is changed to std::basic_string<T, Traits, Allocator> instead of String the caller can simply call make_vector_from_string(bar); without having to specify the template parameter. Why is that?

Just do `template std::vector make_vector_from_string(const String & str)`; — Jarod42, Jan 14 '20 at 10:34

score 2 · Accepted Answer · answered Jan 14 '20 at 10:46

T is non deducible, only String is.

If you want to provide type automatically for T, Traits, Allocator. You have to do it the other way:

template <
    typename String,
    typename T = typename String::value_type,
    typename Traits = typename String::traits_type,
    typename Allocator = typename String::allocator_type,
    typename Vector = std::vector<String>>
Vector make_vector_from_string(const String& str)
{
    //do something with str parameter

    return {str};
}

But

template <typename String>
std::vector<String> make_vector_from_string(const String& str)
{
    //do something with str parameter

    return {str};
}

Seems enough.

Or even

std::vector<std::string> make_vector_from_string(const std::string& str)
{
    return { str};
}

Why is it not possible to use a template alias in function as parameter and to be automatically deduced?

1 Answers1