why this template function does not recognize the lamda's returned type?

Question

This template function does not recognize the lamda's returned type, even specifing it decommenting '->void'.

Why does it happen?

What could I do to circumvent this problem?

#include<iostream>
#include<array>
template<typename T, typename S, size_t SIZE>
void for_each(std::array<T,SIZE>& arr, S(*func)(int&))
{
    for (auto i{0}; i != arr.size(); ++i)
        func(arr[i]);
}
int main()
{
    std::array<int, 5> five_elems{10, 20, 30, 40, 50};
    for_each(five_elems, [](int& ref)/*->void*/{ref *= 2; std::cout << ref << ' '; });
    //for (auto i : five_elems)
    //    i*=2;
    for (const auto i : five_elems)
        std::cout << i << ' ';
}

Answer 1

Your for_each expects a function pointer, but the implicit conversion (from lambda to function pointer) won't be considered in template argument deduction , which causes the calling failing.

You can perform the conversion explicitly:

for_each(five_elems, static_cast<void(*)(int&)>([](int& ref)/*->void*/{ref *= 2; std::cout << ref << ' '; }));

Or

for_each(five_elems, +[](int& ref)/*->void*/{ref *= 2; std::cout << ref << ' '; });

Or just stop using the function pointer parameter. You can add a new type template parameter and then the lambda directly.

template<typename T, size_t SIZE, typename F>
void for_each(std::array<T,SIZE>& arr, F func)
{
    for (auto i{0}; i != arr.size(); ++i)
        func(arr[i]);
}