CODEWITHSUNDEEP
c++templatesc++17type-deduction
Ian Lovely
Ian Lovely

Reputation: 11

Type deduce a template functions return type

I am writing a template function which returns the difference of consecutive elements of a vector using std::minus.

At present I have the following code:

template <typename T>
auto diffs(std::vector<T> &nums)
{
    typedef decltype(std::minus<T>()) diffType;
    std::vector<diffType> differences;
    differences.reserve(nums.size() - 1);
    auto test = std::back_inserter<std::vector<decltype (std::minus<T>())>>;
    std::transform(++begin(nums), end(nums), begin(nums), std::back_inserter(differences), std::minus<T>());

    return differences;
}

and am calling it using 

void test()
{
    auto values = std::vector<int>{ 1,2,4,6,89,1456 };
    auto differences = diffs(values);
}

When I compile this I get:

binary '=': no operator found which takes a right-hand operand of type 'int' (or there is no acceptable conversion)

How do I declare my typedef so that it works properly with my back_inserter?

Upvotes: 1

Views: 75

Answers (1)

Barry
Barry

Reputation: 303127

This:

decltype(std::minus<T>())

Is asking for the type of default constructing a minus<T>, which is just minus<T>. What you'd wanted is what happens when you invoke it:

decltype(std::minus<T>()(nums[0], nums[0]))

Or you could use a type trait instead:

std::invoke_result_t<std::minus<T>, T const&, T const&>

Upvotes: 5

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