How to reverese a c-style string - c++

Question

I would like to reverse a c-style string and came up with the following code.
Play with the code
#include <iostream>
void reverse_c_str(char *c)
{
char *new_c = c;
for (; *new_c; ++new_c){} // find '\0'
--new_c; // new_c is now at the actual last char, which is 'g'
while (c < new_c) // swap chars
{
char tmp = *c;
*c = *new_c; // crash
*new_c = tmp;
--new_c;
++c;
}
}
int main()
{
char *c = "my string";
reverse_c_str(c);
std::cout << c << '\n';
}
Unfortunately, my code has an error, which I marked with // crash. Why does this line crash?

Answer 1

"my string" is a string literal, it is const.
It can be cast to a non-const char * for reasons of compatibility with C, but modifying a constant invokes undefined behavior. In some cases the OS will prevent it (as when it's stored in a read-only section), which is probably what you're observing.
Make a copy of the string in automatic storage, then you'll be able to modify it:
int main()
{
char c[] { "my string" };
reverse_c_str(c);
std::cout << c << '\n';
}

Answer 2

And of course there is a templated soultion:
#include <cstring>
template<std::size_t N>
void reverse_c_str(char (&str)[N]) {
std::size_t len = N-1;
for (std::size_t i = 0; i < len/2; ++i) {
str[i] ^= str[len-i-1];
str[len-i-1] ^= str[i];
str[i] ^= str[len-i-1];
}
}
int main() {
char c[] {"123"};
reverse_c_str(c);
char c2[] {""};
reverse_c_str(c2);
char c3[] {"4321"};
reverse_c_str(c3);
return 0;
}

Answer 3

Use std::swap and std::strlen. I made example for you here.
#include <iostream>
#include <cstring>
void reverse_c_str(char *c) {
int length = std::strlen(c);
for (int i = 0; i < length / 2; i++)
std::swap(c[i], c[length - i - 1]);
}
int main()
{
char c[] { "my string" };
reverse_c_str(c);
std::cout<<c << std::endl;
return 0;
}
Output:
gnirts ym
Another one version of function reverse_c_str
void reverse_c_str(char *c) {
if(*c) {
for(auto begin = c, end = c + std::strlen(c) - 1;
begin < end; ++begin, --end
) {
std::swap(*begin, *end);
}
}
}