How to convert string to char array in C++?

Question

Simplest way I can think of doing it is:

string temp = "cat";
char tab2[1024];
strcpy(tab2, temp.c_str());

For safety, you might prefer:

string temp = "cat";
char tab2[1024];
strncpy(tab2, temp.c_str(), sizeof(tab2));
tab2[sizeof(tab2) - 1] = 0;

or could be in this fashion:

string temp = "cat";
char * tab2 = new char [temp.length()+1];
strcpy (tab2, temp.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.

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.

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());

Just copy the string into the array with strcpy.

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'; 

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