How do I overload ostream::operator<<

Hi, everyone. This code compiles and runs, producing the expected output:

Code:

---

#include <iostream>
 
template<typename T, std::size_t n>
std::ostream& operator<<(std::ostream& os, const T (&arr)[n])
{
    std::size_t i;
 
    for (i = 0; i < n; ++i)
    {
        if (i > 0)
            os << ' ';
 
        os << arr[i];
    }
 
    return os;
}
 
int main()
{
    int n[10] = { 1,2,3,4,5,6,7,8,9,10 };
    std::string s[2] = { "Hello, world", "Goodbye, Earth" };
 
    std::cout << n << '\n';
    std::cout << s << '\n';
}

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My concern is that the line os << ' '; must be given a single character. If I give it a zero-terminated char array (such as " ") then the compiler doesn't know whether to recurse into my user-defined function, or use the regular ostream:: operator<<().

How do I say "use the one that you had before this function existed"? Or even better, "don't use this function if you already have one", or "don't use this one for char arrays".

Don't do this; write a named function (template) for output instead. Prefer to reserve the overloading of operator<< for ostream for class types (or class templates, if applicable) that have a canonical output format. This way, you follow the principle of least surprise.

You could do something like this:

Code:

---

#include <iostream>
#include <string>

template<typename T, std::size_t N>
struct arrayout_obj {
    const T (&arr)[N];
    const char *sep;
    arrayout_obj(const T (&arr)[N], const char *sep) : arr(arr), sep(sep) {}
};

template<typename T, std::size_t N>
auto arrayout(const T (&arr)[N], const char *sep = ", ") {
    return arrayout_obj<T,N>(arr, sep);
}

template<typename T, std::size_t N>
std::ostream& operator<<(std::ostream& os, const arrayout_obj<T,N>& ap) {
    for (std::size_t i = 0; i < N; ++i) {
        if (i > 0) os << ap.sep;
        os << ap.arr[i];
    }
    return os;
}

int main() {
    int n[] = { 1,2,3,4,5,6,7,8,9,10 };
    std::string s[] = { "Mercury", "Venus", "Earth", "Mars" };
    std::cout << arrayout(n) << '\n';
    std::cout << arrayout(s, "\n");
}

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