Does >> leave simbolst in the stream?

Question

The code I wrote below figures the sum of the digits the given number consists of until sum > 10. So,the questions is: operator ">>" extracts the information from the stream and makes the stream empty ? If it's ,so why can't I perfom "ss<< sum" after I reset the bit of EOF to 0 ?

int sum = 0;
stringstream ss("13245");
char ch;
while (1)
{
    while (ss >> ch)
    {
        sum += ch - '0';
    }
    ss.clear();
    ss << sum; //can't perfom
    if (sum<10 ) break;
             sum=0;


}
cout << sum;

Answer 1

If you're trying to reset the stream with each iteration, this will likely do what you seek:

int sum = 0;
stringstream ss("13245");

while (1)
{
    int c;
    while ((c = ss.get()) != EOF)
        sum += c - '0';

    ss.clear(); // clear stream state
    ss.str(""); // clear buffer
    ss << sum;  // write new data

    if (sum<10 )
        break;

    sum=0;
}
cout << sum;

Output

6

Note: I took liberty to use the get() member, as it seemed more appropriate for what you were trying to accomplish. And tested with your 9999 aux sample also produces the expected 9 result. To answer your question, yes, the string buffer is not cleared unless you clear it (which we do above with str("")).

Answer 2

Isn't your problem better solved with integer than with strings? Like:

#include <iostream>

template<typename SomeIntegralT>
SomeIntegralT sum_digits(SomeIntegralT n) {
    do {
        SomeIntegralT sum = 0;
        while( n ) {
            sum += n % 10;
            n /= 10;
        }
        n = sum;
    } while( n > 9 );
    return n;
}

int main()
{
    std::cout << sum_digits(124343525ul) << std::endl;
    std::cout << sum_digits(9999) << std::endl;
    std::cout << sum_digits(12345) << std::endl;
}