[c++] How to convert string to char array in C++?

I would like to convert string to char array but not char*. I know how to convert string to char* (by using malloc or the way I posted it in my code) - but that's not what I want. I simply want to convert string to char[size] array. Is it possible?

#include <iostream>
#include <string>
#include <stdio.h>
using namespace std;

int main()
{
    // char to string
    char tab[4];
    tab[0] = 'c';
    tab[1] = 'a';
    tab[2] = 't';
    tab[3] = '\0';
    string tmp(tab);
    cout << tmp << "\n";

    // string to char* - but thats not what I want

    char *c = const_cast<char*>(tmp.c_str());
    cout << c << "\n";

    //string to char
    char tab2[1024];
    // ?

    return 0;
}

If you don't know the size of the string beforehand, you can dynamically allocate an array:

auto tab2 = std::make_unique<char[]>(temp.size() + 1);
std::strcpy(tab2.get(), temp.c_str());

Easiest way to do it would be this

std::string myWord = "myWord";
char myArray[myWord.size()+1];//as 1 char space for null is also required
strcpy(myArray, myWord.c_str());

Ok, i am shocked that no one really gave a good answer, now my turn. There are two cases;

1. A constant char array is good enough for you so you go with,

   const char *array = tmp.c_str();
   

2. Or you need to modify the char array so constant is not ok, then just go with this

   char *array = &tmp[0];
   

Both of them are just assignment operations and most of the time that is just what you need, if you really need a new copy then follow other fellows answers.

Well I know this maybe rather dumb than and simple, but I think it should work:

string n;
cin>> n;
char b[200];
for (int i = 0; i < sizeof(n); i++)
{
    b[i] = n[i];
    cout<< b[i]<< " ";
}

Try this way it should be work.

string line="hello world";
char * data = new char[line.size() + 1];
copy(line.begin(), line.end(), data);
data[line.size()] = '\0'; 

Just copy the string into the array with strcpy.

str.copy(cstr, str.length()+1); // since C++11
cstr[str.copy(cstr, str.length())] = '\0';  // before C++11
cstr[str.copy(cstr, sizeof(cstr)-1)] = '\0';  // before C++11 (safe)

It's a better practice to avoid C in C++, so std::string::copy should be the choice instead of strcpy.

If you're using C++11 or above, I'd suggest using std::snprintf over std::strcpy or std::strncpy because of its safety (i.e., you determine how many characters can be written to your buffer) and because it null-terminates the string for you (so you don't have to worry about it). It would be like this:

#include <string>
#include <cstdio>

std::string tmp = "cat";
char tab2[1024];
std::snprintf(tab2, sizeof(tab2), "%s", tmp.c_str());

In C++17, you have this alternative:

#include <string>
#include <cstdio>
#include <iterator>

std::string tmp = "cat";
char tab2[1024];
std::snprintf(tab2, std::size(tab2), "%s", tmp.c_str());

You could use strcpy(), like so:

strcpy(tab2, tmp.c_str());

Watch out for buffer overflow.

Try strcpy(), but as Fred said, this is C++, not C