Question

See the following code, debug and show convert is successful on both iPhone simulator and device (4S), but I wonder how does it work? See http://www.cplusplus.com/reference/iostream/ostream/operator%3C%3C/, no overload function for boost::int64_t.

Any risk if I use this function to convert any arbitrary boost::int64_t type? Thanks in advance.

std::stringstream mySS;
boost::int64_t large = 4294967296889977;
mySS<<large;
std::string str = mySS.str(); 

Answer 1

The reason it works is that boost:int64_t is really a typedef to a built-in type (typically std::int64_t defined in cstdint or something like that), so it probably ends up being the same as long long (or similar, depending on the platform). Of course there is an overload of stringstream::operator<< for that.

For the exact definition, best see boost/cstdint.hpp (1.51 version).

It is probably a relatively safe bet to assume that this will generally work on all major platforms. But I doubt anyone will be able to give a guarantee for that.

If the purpose of your using std::stringstream is to convert between integers and strings, the safest thing you can do is simply to use Boost’s own way of converting: boost::lexical_cast (1.51 version). Here is how it’s done:

#include <iostream>
#include <boost/cstdint.hpp>
#include <boost/lexical_cast.hpp>

int main()
{
  boost::int64_t i = 12;
  std::string    s = boost::lexical_cast<std::string>(i);

  std::cout << s << std::endl;
  return 0;
}