Use strcpy to transform a C++ string to a Char array

Question

For some reason, I was trying to covert a C++ string to a char array. Below is what I did:

string aS="hello world";
char aC[aS.size()];
strcpy(aC, aS.c_str());
cout << aC[0] << endl;

string bS="veooci  m eode owtwolwwwwtwwj mooawee mdeeme  eeeec eme aemmeledmll llllleolclclcmslococecewaccocmelaeaccoaaooojutmjooooocoemoooealm omjcdmcmkemmdemmmiwecmcmteeeote eoeeeem   ecc e  yolc e w    dtoooojttttmtwtt ttjcttoowl otdooco ko mooooo  aowmemm o  et  jmc    cmlctmmcccjcccecomatocooccoeoclooomoecwooo mcdoo dcdco dddooedoemod eddeedoedje  emadleweemeeedeeeec   or o  m  wejeetoj    o ojjjlwdjjjj mjmceaeoaai laaadoaa aetmotaemmj  mmmmmmlmm cmol c  mwoaoe omav";

char bC[bS.size()];
strcpy(bC, bS.c_str());
cout << aC[0] << endl;

When aC[0] was first called, it gave 'h' as expected. However, when aC[0] is called at the second time, it gave ''. Could someone explain me what happened here?

`char aC[aS.size()];` is not standard C++ (array dimensions need to be constant expressions, you could use `new` instead). — crashmstr, Apr 07 '15 at 16:55

score 6 · Answer 1 · answered Apr 07 '15 at 16:54

6

For some reason, I was trying to covert a C++ string to a char array

Why don't you just use .c_str(), it does this for you

string aS="hello world";
cout << aS.c_str();

answered Apr 07 '15 at 16:54

Cory Kramer

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Paul R · Answer 2 · 2015-04-07T17:04:58.607

4

char aC[aS.size()];

would need to be:

char aC[aS.size() + 1];

in order to allow for the terminating '\0' required for a C string.

edited Apr 07 '15 at 17:04

answered Apr 07 '15 at 16:57

Paul R

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score 1 · Answer 3 · answered Apr 07 '15 at 17:01

1

If you don't want to use the std::string::c_str() member function, you should then allocate the char array dynamically, as you cannot declare variable length arrays in C++, such as:

char* bC = new char[aS.size() + 1]; // don't forget to delete[]
strcpy(bC, bS.c_str());

answered Apr 07 '15 at 17:01

vsoftco

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May I know how to get the size of bC? Since it is a pointer, sizeof() does not work. – Just don't rain Apr 07 '15 at 17:36
@Justdon'train I thought you want to "copy" the string into `bC`. Why do you need to get its size? – vsoftco Apr 07 '15 at 17:37
I wanna to do some stuff for each character. – Just don't rain Apr 07 '15 at 17:39
@Justdon'train well the size of it is simply the size of the string you copied, i.e. `aS.size() + 1 `. You can also use `strlen(bC)` to get the size, since it is a C-string. – vsoftco Apr 07 '15 at 17:40
yes. just for curiosity, if we are given a pointer that points to an array, is there a way to get its size? – Just don't rain Apr 07 '15 at 17:42
@Justdon'train The size of the array? No, unless there is some special delimiter (such as '\0' in case of C-strings) – vsoftco Apr 07 '15 at 17:43

PaulMcKenzie · Answer 4 · 2015-04-07T17:32:54.310

I can see some utility of changing to an array if you're calling a function that will change the string buffer (pre C++11). Otherwise, use the other solutions such as c_str() if the buffer is not going to be changed.

However, if you really wanted a char array due to the buffer possibly being changed, use std::vector<char>. This will bypass the issue of using non-standard syntax (aC[aS.size()]), and removes the need to dynamically allocate memory. A std::vector<char> effectively is a changeable string buffer, only wrapped.

#include <vector>
#include <string>
#include <cstring>
#include <iostream>

using namespace std;

int main()
{
    string aS="hello world";
    std::vector<char> aC(aS.begin(), aS.end());  // copy
    aC.push_back(0);  // null-terminate
    cout << &aC[0] << endl;
}

Output:

hello world

Now, where you need to specify a character array pointer, you just pass the address of the first element in the vector (provided that the vector is not empty).

Live Example: http://ideone.com/tF32Zt

Use strcpy to transform a C++ string to a Char array

4 Answers4