C++ – const char * const versus const char *

I'm running through some example programs to refamiliarize myself with C++ and I have run into the following question. First, here is the example code:

void print_string(const char * the_string)
{
    cout << the_string << endl;
}

int main () {
    print_string("What's up?");
}

In the above code, the parameter to print_string could have instead been const char * const the_string. Which would be more correct for this?

I understand that the difference is that one is a pointer to a constant character, while the other one is a constant pointer to a constant character. But why do both of these work? When would it be relevant?

Best Answer

The latter prevents you from modifying the_string inside print_string. It would actually be appropriate here, but perhaps the verbosity put off the developer.

char* the_string : I can change which char the_string points to, and I can modify the char to which it points.

const char* the_string : I can change which char the_string points to, but I cannot modify the char to which it points.

char* const the_string : I cannot change which char the_string points to, but I can modify the char to which it points.

const char* const the_string : I cannot change which char the_string points to, nor can I modify the char to which it points.

Related Solutions

Why can’t I convert ‘char**’ to a ‘const char* const*’ in C

I had this same problem a few years ago and it irked me to no end.

The rules in C are more simply stated (i.e. they don't list exceptions like converting char** to const char*const*). Consequenlty, it's just not allowed. With the C++ standard, they included more rules to allow cases like this.

In the end, it's just a problem in the C standard. I hope the next standard (or technical report) will address this.

C++ – How to convert a std::string to const char* or char*

If you just want to pass a std::string to a function that needs const char* you can use

std::string str;
const char * c = str.c_str();

If you want to get a writable copy, like char *, you can do that with this:

std::string str;
char * writable = new char[str.size() + 1];
std::copy(str.begin(), str.end(), writable);
writable[str.size()] = '\0'; // don't forget the terminating 0

// don't forget to free the string after finished using it
delete[] writable;

Edit: Notice that the above is not exception safe. If anything between the new call and the delete call throws, you will leak memory, as nothing will call delete for you automatically. There are two immediate ways to solve this.

boost::scoped_array

boost::scoped_array will delete the memory for you upon going out of scope:

std::string str;
boost::scoped_array<char> writable(new char[str.size() + 1]);
std::copy(str.begin(), str.end(), writable.get());
writable[str.size()] = '\0'; // don't forget the terminating 0

// get the char* using writable.get()

// memory is automatically freed if the smart pointer goes 
// out of scope

std::vector

This is the standard way (does not require any external library). You use std::vector, which completely manages the memory for you.

std::string str;
std::vector<char> writable(str.begin(), str.end());
writable.push_back('\0');

// get the char* using &writable[0] or &*writable.begin()