Why is the template found in the constraint but not the function? C++20

Question

I'm learning about C++20 concepts and constraints and I don't understand why in this case T is considered undefined, yet it seems to work fine in the requires clause. What's wrong here? and why?

If I remove the semicolon, the compiler says U is not a template

error: expected unqualified-id before ‘;’ token
   11 | requires std::integral<T> || std::floating_point<T>;
      |                                                    ^
main.cpp:12:28: error: ‘T’ was not declared in this scope
   12 | constexpr double average(U<T> const &it) {

#include <numeric>
#include <vector>
#include <iostream>
#include <concepts>

template <typename T, std::forward_iterator U>
requires std::integral<T> || std::floating_point<T>;
constexpr double average(U<T> const &it) {
    const double sum = std::accumulate(it.begin(), it.end(), 0.0);
    return sum / it.size();
}

int main() {
    std::vector ints { 1, 2, 3, 4, 5};
    std::cout << average(ints) << '\n';
}

The compiler is confused by the meaning of the `;`. The stuff after that is irrelevant. Also "U is not a template" is correct; it's not a template. It's a typename. But you tried to use it as a template when you said `U`. That's the problem. — Nicol Bolas, Jun 07 '22 at 03:49
@康桓瑋 I'm trying to write a function that calculates the average of an iterator U having a type T which is integral or float. — Cedric Martens, Jun 07 '22 at 04:44
You want the function to receive an iterator, but you're passing in a `vector`? — 康桓瑋, Jun 07 '22 at 05:53
@CedricMartens: Iterators do not "have a type T". Iterators have a `value_type`, but you don't say that by saying `IteratorType` — Nicol Bolas, Jun 07 '22 at 06:18

score 0 · Answer 1 · answered Jun 07 '22 at 11:21

please read about container concept for the full solution. here is a partial solution just to show a direction.

#include <numeric>
#include <vector>
#include <iostream>
#include <concepts>

template <typename T>
requires requires (std::integral<typename T::value_type>) || (std::floating_point<typename T::value_type>)
constexpr double average(T const &container) {
    const double sum = std::accumulate(container.begin(), container.end(), 0.0);
    return sum / container.size();
}

int main() {
    std::vector ints { 1, 2, 3, 4, 5};
    std::cout << average(ints) << '\n';
}

Why is the template found in the constraint but not the function? C++20

1 Answers1