Why is `“literal”` encouraged to decay to `const char*` in C++ argument type match?

Question

I\'m playing around with overloading operators in c++14, and I tried to match two types of arguments: any-old-const-char*, and a-string-literal.

That is, I\'m trying

Answer 1

Why is const char* preferred over const char (&)[N]?

The reason for this is rather technical. Even though the decay of a string literal from const char[N] to const char* is a conversion, it falls into the "lvalue transformation" category and is therefore considered by [over.ics.rank]/3 to be as good as no conversion at all. Since "no conversion" is required for either overload, the non-template overload wins.

Why is const char (&)[N] preferred over const char (&)[] (non-template)?

It is not possible to bind a reference to array of unknown bound to a value of type array of known bound. Instead, a reference to array of unknown bound can only be bound to values that are themselves arrays of unknown bound.

Why is const char (&&)[N] unable to compile?

A string literal is an lvalue so I'm not sure why you would expect this to work.

Is there a "right way" to capture literal strings?

You can use a helper function template that captures its argument using a forwarding reference so as to not destroy any type information (const char* versus const char[N]) then dispatch on the type using template specialization. You'll probably also want to use SFINAE to make sure it is disabled if anything other than a const char* or const char[N] is passed in. To wit,

template 
struct f_helper;

template <>
struct f_helper {
    void do_it(const char*) {
        puts("pointer");
    }
};

template <>
struct f_helper {
    template 
    void do_it(const char (&)[N]) {
        printf("array of length %zd\n", N);
    }
};

template >::value ||
                                                   std::is_same>::value>::type>
void f(T&& s) {
    f_helper>::value>{}.do_it(s);
}

Coliru link: http://coliru.stacked-crooked.com/a/0e9681868d715e87