#include <iostream>
#include <sstream>

int main() {
    int i = 10;

    using namespace std;
    stringstream str;
    str << i;
    string s = str.str();
    cout << s << endl;
    return 0;
}

char* itoa(int val, int base) 
{
    static char buf[32] = {0};
    int i = 30;
    for(; val && i ; --i, val /= base)
            buf[i] = "0123456789abcdef"[val % base];
    return &buf[i+1];
}

Note Each function uses a static variable for parsing the string into tokens. If multiple or simultaneous calls are made to the same function, a high potential for data corruption and inaccurate results exists. Therefore, do not attempt to call the same function simultaneously for different strings and be aware of calling one of these functions from within a loop where another routine may be called that uses the same function. However, calling this function simultaneously from multiple threads does not have undesirable effects.

CONFORMING TO
C89 describes strtod(), C99 describes the other two functions.

  class itoa 
  {
    char buffer[38];
  public:
    itoa(int n, int radix = 10) { ::itoa(n,buffer,radix); }
    operator const char*() { return buffer; }
  };

std::string s = itoa(123,16);

CS>>  class itoa 
CS>>  {
CS>>    char buffer[38];
CS>>  public:
CS>>    itoa(int n, int radix = 10) { ::itoa(n,buffer,radix); }
CS>>    operator const char*() { return buffer; }
CS>>  };
CS>>

   char buffer[ sizeof(int) * 8 + 1 ];

N>А вообще подход интересный — такого рода класс позволяет делать кэширование результатов.

char *itoa(int n, char *s, int b) 
{
        if(b < 2 || b > 36) return s; 

    static char digits[] = "0123456789abcdefghijklmnopqrstuvwxyz";
    int i=0, sign;
    
    if ((sign = n) < 0)
      n = -n;

    do s[i++] = digits[n % b];
    while ((n /= b) > 0);

    if (sign < 0)
      s[i++] = '-';
    s[i] = '\0';
    return strrev(s);
}

N>>>А вообще подход интересный — такого рода класс позволяет делать кэширование результатов.