C++ converting an int to an array of chars?

Question

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C++ convert int and string to char*

Hello, i am making a game and I have a score board in it. The score is stored in an int variable but the library im using for the game needs an array of chars for its text outputting for my scoreboard.

So how do i turn an int into an array of chars?

int score = 1234;  // this stores the current score

dbText( 100,100, need_the_score_here_but_has_to_be_a_char_array); 
// this function takes in X, Y cords and the text to output via a char array

The library im using is DarkGDK.

tyvm :)

Answer 1

char str[10];  
sprintf(str,"%d",value);

Answer 2

ostringstream sout;
sout << score;
dbText(100,100, sout.str().c_str());

Answer 3

You can use an std::ostringstream to convert the int to an std::string, then use std::string::c_str() to pass the string as a char array to your function.

Answer 4

char str[16];
sprintf(str,"%d",score);
dbText( 100, 100, str );

Answer 5

Well, if you want to avoid C standard library functions (snprintf, etc.), you could create a std::string in the usual manner (std::stringstream, etc.) and then use string::c_str() to get a char * which you can pass to the library call.

Answer 6

Use sprintf

#include <stdio.h>

int main () {
  int score = 1234; // this stores the current score
  char buffer [50];
  sprintf (buffer, "%d", score);
  dbText( 100,100,buffer);

}

Answer 7

Let me know if this helps.

#include <iostream>
#include <stdlib.h>
using namespace std;

int main() {
    char ch[10];
    int i = 1234;
    itoa(i, ch, 10);
    cout << ch[0]<<ch[1]<<ch[2]<<ch[3] << endl; // access one char at a time
    cout << ch << endl; // print the whole thing
}