How to pass a template type (std::array or std::vector) to a template parameter

Question

I have a function template where the return type of the function is the template argument. According to https://en.cppreference.com/w/cpp/language/template_argument_deduction, we can have a template argument deduction with the following examples

template<typename To, typename From>
To convert(From f);

Now as an example, I want to make a specialized function for this where I passed the following:

auto x = convert<std::array>(6000); // the From should be deduced as int.

In this case, I want to make a specialized function template, where the sizeof the From determines the size of the array , and to keep it simple, assume that uint8_t is the type of the array as the data type which represents 1 byte.

Therefore, the type of x should be std::array<uint8_t, 4>.

Another example:

auto y = convert<std::array>(1.02322); // the From should be deduced as double

In this case, the type of y should be std::array<uint8_t, 8>.

The second problem is if std::vector is passed as a template argument

auto x = convert<std::vector>(6000); // the From should be deduced as int

The type of x should be std::vector<uint8_t>, and the length of the vector is 4

How to make such function template that can accept both std::array and std::vector, and without using any std::span algorithm as I am limited to c++11/14

Caleth · Accepted Answer · 2023-01-06 10:00:06Z

3

std::array and std::vector are different kinds of template.

To match std::vector, you can have

template <template <typename...> class To, typename From>
To<uint8_t> convert(From from);

To match std::array, you need

template <template <typename, std::size_t> class To, typename From>
To<uint8_t, sizeof(From)> convert(From from);

But this is good, because you can't partially specialise a function template. Here you instead have overloads.

See it on coliru

edited Jan 6, 2023 at 10:00

answered Jan 6, 2023 at 9:54

Caleth

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