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

Question

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円';
 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;
}

• cegprakash: possible duplicate of std::string to char*

Answer 1

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());

• Fred Larson: The catch with strncpy is that it won't null-terminate if it reaches the size before it finds a null in the source string. Gotta be careful with that too!
• Fred Larson: Yes, I think that's a reasonable way to handle it.
• bames53: strncpy is for safety in that strncpy will not overrun the buffer you give it. Also strcpy_s is not in C99, but it has recently been added in C11.
• Karoly Horvath: I love the way people abuse the word safety... oh, there's not enough room for the string, so let's truncate it.. oh, we accidentally removed the info about the patient's life threatening drug allergy.. but we don't have a buffer overrun anymore. well, I guess it's safe...
• Chowlett: @KarolyHorvath - safety in different domains. Obviously, you need to check your functional requirements and Not Do Stupid Things. But if you overrun the buffer you lose any guarantee of knowing what will happen. Correct coding ensures the program is "safe" to run on your OS. Correct thinking ensures the program is "safe" at doing what the user needs.

Answer 2

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.

• harogaston: He wants a char array, not a char *
• JustinCB: The pointer he created is the same thing as a char array. An array variable in C and C++ is just a pointer to the first element in the array.
• aniliitb10: @JustinC.B. not really! array variables might decay into pointers but they are not same in C++. e.g. std::size(elem) in C++ is well defined when elem is a char array but it fails to compile when elem is a char* or const char*, you can see it at here
• William Bradley: Will #2 cause problems (dangling pointer) if tmp is stack-allocated and goes out of scope but array has a longer lifetime?
• Bersan: @WilliamBradley yes, you will access invalid memory

Answer 3

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

Answer 4

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());

• interjay: Array size must be a compile-time constant in C++.
• interjay: @PhaniRithvij It's wrong unless you're only targeting a compiler that allows this as a language extension, in which case it's still bad practice.

Answer 5

Just copy the string into the array with strcpy.

Answer 6

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円'; 

Answer 7

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

Answer 8

You could use strcpy(), like so:

strcpy(tab2, tmp.c_str());

Watch out for buffer overflow.

Answer 9

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());

Answer 10

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());

Answer 11

Well I know this maybe rather dumb (削除) than (削除ここまで) 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]<< " ";
}

• droidballoon: Well, not really. You're just assuming that n can't be larger than b. It would crash if n>b. Better to use the strcpy function already provided. Edit: At some cases it might be even better to use strcpy_s as it adds check for NULL pointers (if your platform supports it)