Question

void f(char  *  & pch) {
    cout << &pch << " " << pch << endl;
}

int main() {
    char *pch2 = new char[11];
    strcpy(pch2, "1234567890");
    cout << &pch2 << " " << pch2 << endl;
    f(pch2);

    return 0;
}

gives next output:

0xbfa0d62c 1234567890 
0xbfa0d62c 1234567890

but if i would modify first row as follows

void f(char const * const & pch) {

i will get:

0xbfec7df8 1234567890
0xbfec7dfc 1234567890

is difference in pointers appeared because of need to mark new memory cell as const or something else?

Answer 1

pch2 is a char*, not a char const*. You cannot bind a reference of type char const*& to a pointer of type char*, so the following would be ill-formed:

char* p(0);
char const*& r(p);

Similarly, if your function was declared as void f(char const*& pch), you would not be able to call it with a char* argument because of the const-qualifier mismatch.

The reason that your example works is that a const reference can bind to a temporary, and the compiler can construct a temporary copy of your char* pointer, give that temporary the type char const*, and bind the reference pch to that temporary pointer.