C++ print template container error (error: ambiguous overload for 'operator<<') understanding?

Question

I want to write template function which can print container like std::vector, std::list.

Below is my function, just overload <<.

Answer 1

@miradham explained the issue very well.

but this a more generic solution using SFINAE to make the overload considered only for types on which the range-based for loop can be used, what ever their template arguments can be.

type coming form std::basic_string were ignored to prevent ambiguity with the standard operator << to display strings

c-style array will not be displayed using this overload even though they could because they are decayed to pointers and displayed with the standard operator <<

#include 
#include 
#include 
#include 
#include 

template typename From, typename T>
struct is_from : std::false_type {};

template typename From, typename ... Ts>
struct is_from > : std::true_type {};

template 
using void_t = void;

template 
struct is_input_iterator : std::false_type { };

template 
struct is_input_iterator()),
           decltype(*std::declval()),
           decltype(std::declval() == std::declval())>>
    : std::true_type { };

template()))>::value &&
                        is_input_iterator()))>::value &&
                        !is_from::value, int>::type = 0>
std::ostream& operator<<(std::ostream& out, const Container& c){
    for(const auto& item:c){
        out << item << " ";
    }
    return out;
}

int main(){

    std::array arr{0, 1, 2, 3, 4, 5};
    std::vector vec{5, 9, 1, 4, 6};

    std::cout << vec << std::endl;
    std::cout << arr << std::endl;
    std::cout << std::string("test") << std::endl;
    return 0;
}