Re: returning reference to pointer

Hi All

I'm hoping someone can explain this to me. If I compile the following
code segment:

#include <iostream>
#include <cstring>

char const* max (char const* a, char const* b)
{
     return std::strcmp(a,b) < 0 ? b : a;

}

template <typename T>
T const& max (T const& a, T const& b, T const& c)
{
     return max (max(a,b), c);

}

int main ()
{
     const char* s1 = "foo";
     const char* s2 = "bar";
     const char* s3 = "betty";
     std::cout << "::max(s1, s2, s3) = " << ::max(s1, s2, s3) <<
std::endl;

}

I get "warning: returning reference to temporary" from gcc 4.0.1.
Fine, I understand returning a reference to a temporary is bad for
obvious reasons. But out of curiousity I changed the shape of the 2
parameter max function definition to:

char const*& max (char const* a, char const* b)

i.e. to return a reference to a pointer and now it compiles and the
program runs printing the correct value.

Does this mean the modified function now returns a reference to what
the copied pointer points to?