Using char* to store integers - c++

Question

I am trying to convert an integer into a string of numerals stored in char *s.
char* ItoA(int n, char *s){
int mod = n;
for(int x = 1; x <= sizeof(n) + 2; x++){
int digit = mod % 10;
s[x-1] = digit;
mod = mod / 10;
}
return s;
}
void main(){
int n = 12345;
char s3;
// should print 12345
cout << "\n" << ItoA(n, &s3);
cin.get();
}
I know that something is wrong with my notation because I keep getting an array of garbage as the output in main. Why am I getting garbage instead of "12345"?

Answer 1

Q: Why am I getting garbage instead of "12345"?
You are using &s3 as though it were an array of characters.
Use:
int main(){
int n = 12345;
char s3[20]; // Use an array of characters large enough to hold the number.
// should print 12345
cout << "\n" << ItoA(n, s3);
cin.get();
}
Also, it's not clear to me what your logic is for using sizeof(n) below:
for(int x = 1; x <= sizeof(n) + 2; x++)

Answer 2

In order to convert a number to a string, you need to convert to ASCII by adding the character '0'. Also, you need to allocate enough memory to store all the characters and initialize the memory.
Finally the logic of your for loop is broken, and it is generally easier to construct the string backwards and then reverse it.
#include <iostream>
#include <string.h>
using namespace std;
char* ItoA(int n, char *s, int bufLength){
// Create the string in reverse order
int i;
for(i = 0; i < bufLength; i++){
int digit = n % 10;
s[i] = digit + '0';
n /= 10;
if (n == 0) break;
}
// Reverse the string.
for (int j = 0; j <= i / 2; j++) {
char c = s[j];
s[j] = s[i - j];
s[i - j] = c;
}
return s;
}
int main(){
int n = 12345;
char s3[6];
memset(s3, 0, 6);
// should print 12345
cout << "\n" << ItoA(n, s3, 6);
cin.get();
}