overloading std::to_string and std::to_chars for a custom type?

Question

I have an arithmetic custom type with a to_string() member. I'm thinking about overloading std::to_string in the std namespace for the type, but would this count as namespace pollution?

EDIT: I'm thinking about this, so that this expression SFINAE would work for the type: -> decltype(std::to_string(std::declval<T>()), void(0))

[This extending the `std` namespace reference](https://en.cppreference.com/w/cpp/language/extending_std) could be useful. In short, overloading e.g. `std::to_string` is not allowed (and leads to UB). — Some programmer dude, Jul 07 '18 at 06:21
An alternative could be to keep `to_string` and `to_chars` in your own namespace, and than [leverage argument dependent lookup](https://stackoverflow.com/q/4782692/2486888). But it may be tricky. — , Jul 07 '18 at 06:28

score 3 · Answer 1 · answered Jul 07 '18 at 07:24

As Some programmer dude pointed out in the comment, you cannot add declarations into std namespace in this case.

Like this post, a workaround is to write your to_string in the same namespace as your custom class, and use using std::to_string to bring std::to_string in overload resolution when you want to use to_string for generic type. The following is an example:

#include <string>

struct A {};

std::string to_string(A) {return "";}

namespace detail { // does not expose "using std::to_string"
    using std::to_string;
    template <typename T>
    decltype(to_string(std::declval<T>()), void(0)) foo() {}
}
using detail::foo;

int main()
{
    foo<int>();      // ok
    foo<A>();        // ok
    // to_string(0); // error, std::to_string is not exposed
}

overloading std::to_string and std::to_chars for a custom type?

1 Answers1