Question

When I put something into a stringstream, let’s say a real number, if I then insert that stringstream object into cout…what am I looking at?

Usually I’m getting some strange number. Is this a memory location? Just curious.

It looks like the below comment hit it but here’s what I’m trying to do:

string stringIn; 
stringstream holdBuff;
holdBuff << getline(cin, stringIn);
cout << holdBuff; 

Basically I was just trying to see what holdBuff looked like once I inserted stringIn. I am trying to have the user enter a string and then I want to step through it looking for it’s contents and possilbly converting…

Answer 1

What do you think

holdBuff << getline(cin, stringIn);

is doing. The return type of getline is a reference to the stream
being read (cin) in this case. Since there’s no << defined which
takes an std::istream as second argument, the compiler tries different
conversions: in C++11, std::istream has an implicit conversion to
bool, and in earlier C++, an implicit conversion to std::ios*, or
something similar (but the only valid use of the returned value is to
convert it to bool). So you’ll either output 1 (C++11), or some
random address (in practice, usually the address of the stream, but this
is not guaranteed). If you want to get the results of a call to
getline into an std::ostringstream, you need two operations (with a
check for errors between them):

if ( !getline( std::cin, stringIn ) )
    //  Error handling here...
holdBuff << stringIn;

Similarly, to write the contents of a std::ostringstream,

std::cout << holdBuf.str() ;

is the correct solution. If you insist on using an std::stringstream
when an std::ostringstream would be more appropriate, you can also do:

std::cout << holdBuf.rdbuf();

The first solution is preferable, however, as it is far more idiomatic.

In any case, once again, there is no << operator that takes any
iostream type, so you end up with the results of the implicit
conversion to bool or a pointer.