I am reading C++ Primer (5th Edition), 16.5, Defining a Function Template Specialization, and confused about the example given by author, let's see the following template function:
template <typename T> int compare(const T&, const T&);
and its specialization version:
template <>
int compare(const char* const &p1, const char* const &p2)
{
return strcmp(p1, p2);
}
Type of T will be const char *, but I don't think the function could be the specialization version of the template function, because I think const char* const &p1 can only be the specialization of T const & but const T &, I know I'm wrong, but I want to know why I am wrong.
Edit:
One thing to emphasize, if I invoke compare("hi", "mom"), it won't compile, that is to say, template <typename T> int compare(const T&, const T&) can not be initialized to int compare(const char* const &p1, const char* const &p2), I know T will be initialized with char[3] or char[4], but now that this won't compile, why won't compiler ignore this kind of initialization but choose one that will compile?
In C++, const T and T const mean exactly the same thing. Therefore const T & and T const & mean exactly the same thing.
It couldn't really be any other way, because references can never be changed to refer to something else (they are not "reseatable"). If you read T const & as "a reference which cannot be changed to refer to a different T", that's incorrect. It is "a reference to a T, which cannot be used to modify that T (and as with all references, cannot be changed to refer to a different T)."