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

1. The overload taking a pointer is preffered because it is not a template according to

13.3.3 Best viable function [over.match.best]

...

Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then

...

(1.7) F1 is not a function template specialization and F2 is a function template specialization

2. actually non-template const char (&)[] does not seem to compile at all because it is a reference to a non-bound array. It is possible to pass a pointer like this const char [], but not array.

3. this should fail at least for the same reason as (2)

4. you can provide another template taking a reference to pointer:

template< typename = void > void
foo(char const * & text)
{
    ::std::cout << "got ptr" << ::std::endl;
}

Note that providing another template taking a pointer won't work because both template specializations will be fine and we'll get ambiguous choice of overloaded functions.