c++ - std::cout not printing the value with out endl (newline specifier) -

below c++ program multiply 2 strings (integers in strings) , produce integer result in string. believe due cout flush problem. can explain how print value without endl or text string before printing answer.

 #include <iostream>  using namespace std;  string multiply (string s1, string s2)  {     char str[10];     string ans="";     int m=s1.length();     int n=s2.length();      if (!s1.compare("0") || !s2.compare("0"))          return "0";         int *res = new int[m + n];         (int = m - 1; >= 0; i--)        {          (int j = n - 1; j >= 0; j--)          {           res[m + n - - j - 2] += (s1[i] - '0') * (s2[j] - '0');           res[m + n - - j - 1] += res[m + n - - j - 2] / 10;           res[m + n - - j - 2] %= 10;          }        }          (int = m + n - 1; >= 0; i--)        {             if (res[i] != 0)              {                 (int j = i; j >= 0; j--)                  {                   sprintf(str,"%d", res[j]);                   ans+=str;                 }                 return ans;             }          }  }  int main() {   cout << multiply("0", "0"); // doesn't work - prints nothing.  /**cout << multiply("0", "0") << endl; //works!! prints "0" correctly **/   return 0; } 

you should notice std::endl includes flushing output stream. reference documentation:

inserts newline character output sequence os , flushes if calling os.put(os.widen('\n')) followed os.flush().

so if flush cout after printing in example

cout << multiply("0", "0"); cout.put(cout.widen('\n')); cout.flush(); 

you see result printed live demo.

for more information flushing , means buffered output read here please std::ostream::flush().