Pointer to reference

Question

I was reading, and I saw the following code:

template <> inline bool is_empty(const char* const& x) {     return x==0 || *x==0; } 

What does const char* const& x mean?

I tried the following code to understand it:

void f(const char* const& x)  {     // whatever }  void main() {     char a = 'z';     char &j = a;     f(a);//error     f(*a);//error     f(&a);//fine     f(j);//error     f(&j);//fine          f('z');//error } 

It works only for f(&a) and f(&j).

What does const char* const& x actually mean?

Answer 1

cdecl is an useful online tool to help beginners get used to complicated declarations.

Inserting const char* const& i returns:

declare i as reference to const pointer to const char

The meaning of the declaration should be obvious now.

Answer 2

Let's take this apart:

• the trailing & means this is a reference to whatever the type is.

• const char is the type that's being pointed to

• * const means the pointer is constant

So, this is a reference to a const pointer to a const char. You can't change the char that it points to (unless you cast away the constness), and you can't change the pointer (i.e. make it point to something else). The pointer is passed by reference, so that no copying occurs.