const overload functions taking char pointers

12:11:00 PM 0 Comments A+ a-

Always have two overloaded versions of functions that take
char * and const char * parameters. Declare (but don't define if not needed)
a function that takes const char* as a parameter when you have defined
a function that accepts a non-const char* as a parameter.

#include <iostream>
#include <iomanip>

static void foo (char *s) {
std::cout << "non-const " << std::hex << static_cast <void *>(s) << std::endl;
}
static void foo (char const *s) {
std::cout << "const " << std::hex << static_cast <void const *>(s) << std::endl;
}

int main (void)
{
char * c1 = "Literal String 1";
char const * c2 = "Literal String 1";
foo (c1);
foo (c2);
foo ("Literal String 1");
//*c1 = 'l'; // This will cause a seg-fault on Linux.
std::cout << c2 << std::endl;

return 0;
}

Because of default conversion rule from string literal to char *,
the call to foo using in-place literal goes completely undetected
through the eyes of compiler's type system.

Interestingly enough, the addresses of all the identical string literals
is the same, irrespective of whether it is assigned to const or non-const.
Internally though, they are stored on the const DATA page and modifying
them causes a seg-fault.