Can a temporary stringstream object be used?

This doesn't work because the second << is std::ostream &operator<<(std::ostream &, int); and so the return type is ostream& which has no member str().

You would have to write:

result_stream << transform( static_cast<stringstream &>(stringstream() << i).str() );

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Update (2019): According to LWG 1203 the standard may be changed in future (and one major implementation already has) so that this code no longer works, and a simpler code works instead. See this question for detail.

In the interim period, apparently the following works on both old and new:

result_stream << transform( static_cast<stringstream &>(stringstream().flush() << i).str() );
//                                                                    ^^^^^^^^

This should not be a performance penalty since flushing an empty stream has no effect...

The result of the << operator on the temporary stringstream is an ostream. There is no str() method on an ostream.

Use to_string instead:

using std::to_string;

result_stream << transform(to_string(i));

You can define a helper to_string to handle objects not covered by std::to_string.

template <typename T>
std::string to_string (const T &t) {
   std::ostringstream oss;
   oss << t;
   return oss.str();
}

For example, if you had a class Foo that understood redirection to an ostream, and f was an instance of Foo, then you could do:

result_stream << transform(to_string(f));

Try it online!

If you actually want to use a lot of redirection to build up a string before transforming, you could create a helper object for that as well.

struct to_string_stream {
    std::ostringstream oss;
    template <typename T>
    auto & operator << (const T &t) { oss << t; return *this; }
    operator std::string () const { return oss.str(); }
    void clear () { oss.string(std::string()); }
};

Then, you could do something like:

to_string_stream tss;
result_stream << transform(tss << i << ':' << f);

Try it online!

operator<<() returns a reference to the base class std::ostream contained within the std::stringstream. The base class doesn't contain the str() method. You can cast it back down to a std::stringstream&:

result_stream << transform(static_cast<std::stringstream&>(std::stringstream() << i).str());