Thread: transform int to char

use itoa()

Code:

int i;
char c;
cin>>i;
c = itoa(i);

>c = itoa(i);
itoa (usually, but not certainly because it isn't standard) returns a pointer to char. Since c isn't a pointer to char, this won't work too well.

>any ideas how i can make it possible ?
Simple, just use cin's type checking to break the loop for you:

Code:

#include <iostream>

using namespace std;

int main()
{
  int num;

  cout<<"Enter a set of numbers (x to quit): ";
  while ( cin>> num )
    cout<<"You entered "<< num <<'\n';
  cout<<"Had enough, huh?"<<endl;
}

Because num is an integer, cin expects an integer. If it reads a character then it fails and the loop terminates. Of course, if you have any input from cin after the loop you will need to clear the error and discard the unread characters.

>>itoa (usually, but not certainly because it isn't standard)
Really? I thought itoa was standard, but atoi wasn't for some reason.
I've always used it as returning an actual char on the cygwin gcc.
Guess you (meaning I) learn something new every day.