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; }
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11 Answers

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());
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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.

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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());
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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

You could use strcpy(), like so:

strcpy(tab2, tmp.c_str());

Watch out for buffer overflow.

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());

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());

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]<< " "; }
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