CODEWITHSUNDEEP
c++stringstream
MCP
MCP

Reputation: 4536

cout << stringstream

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...

Upvotes: 33

Views: 90113

Answers (4)

Michael Krelin - hacker
Michael Krelin - hacker

Reputation: 143081

cout << s.rdbuf();

is what you want. Alternatively you may want to

cout << s.str();

which may be more expensive in terms of resources though.

Upvotes: 6

James Kanze
James Kanze

Reputation: 153919

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.

Upvotes: 28

Goz
Goz

Reputation: 62323

Yes it is most likely a memory location of some form or other. Most likely it is the pointer to the stringstream object itself.

You could confirm this as follows:

std::stringstream ss;
unsigned long long ll = (unsigned long long)&ss;
cout << ll;

That said when you want to cout a stringstream you should use the str() function as follows:

cout << ss.str();

Upvotes: 6

Bo Persson
Bo Persson

Reputation: 92261

Yes, you are likely to see the address of the stringstream.

If you want to display the string it contains, try

cout << stream.str();

Upvotes: 29

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