C++ error when using stream buffers directly with ?: operator

When I try to use a ternary conditional operator (?:) with stream buffer redirection, gcc produces 'synthesized method first required here' error. What is the problem, and how to correct the following program?

#include <fstream>
#include <iostream>

int main(int argc, char* argv[])
{
    using namespace std;
    cout << cin.rdbuf();    //OK
    ofstream("tmp.txt") << cin.rdbuf(); //OK

    int i=1;
    (i > 1 ? ofstream("tmp.txt") : cout) << cin.rdbuf(); //Compilation ERROR. Why?
    return 0;
}

compiled with gcc4.4:

...    
/usr/include/c++/4.4/bits/ios_base.h: In copy constructor ‘std::basic_ios<char,   std::char_traits<char> >::basic_ios(const std::basic_ios<char, std::char_traits<char> >&)’:  
/usr/include/c++/4.4/bits/ios_base.h:790: error: ‘std::ios_base::ios_base(const std::ios_base&)’ is private  
/usr/include/c++/4.4/iosfwd:47: error: within this context  
/usr/include/c++/4.4/iosfwd: In copy constructor ‘std::basic_ostream<char,   std::char_traits<char> >::basic_ostream(const std::basic_ostream<char,   std::char_traits<char> >&)’:  
/usr/include/c++/4.4/iosfwd:56: note: **synthesized method** ‘std::basic_ios<char,   std::char_traits<char> >::basic_ios(const std::basic_ios<char, std::char_traits<char> >&)开发者_如何转开发’   **first required here**   
../item1_1.cpp: In function ‘int main(int, char**)’:  
../item1_1.cpp:12: note: synthesized method ‘std::basic_ostream<char,   std::char_traits<char> >::basic_ostream(const std::basic_ostream<char,   std::char_traits<char> >&)’ first required here   

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This compiled OK with my version of clang, I think that it may be a gcc bug.

From my reading of the standard, cout is an lvalue of type std::ostream and ofstream("tmp.txt") is an rvalue of type std::ofstream.

Neither has any cv-qualifiers and std::ostream is a base class of std::ofstream so the conditional operator is valid an the result is an rvalue and has type std::ostream.

There is no copying of either operand implied.

if E1 and E2 have class type, and the underlying class types are the same or one is a base class of the other: E1 can be converted to match E2 if the class of T2 is the same type as, or a base class of, the class of T1, and the cv-qualification of T2 is the same cv-qualification as, or a greater cv-qualification than, the cv-qualification of T1. If the conversion is applied, E1 is changed to an rvalue of type T2 that still refers to the original source class object (or the appropriate subobject thereof). [Note: that is, no copy is made. ]

The operator<< overload that you are using is a member of std::ostream so there is no need to bind a temporary to a non-const reference, the member can be called on a non-const rvalue.

basic_ostream<charT,traits>&
    basic_ostream<charT,traits>::operator<< (basic_streambuf<charT,traits>* sb);

Edit

Note that this has changed in C++0x. Now, if the result of a conditional expression is an rvalue a temporary copy is always made. As objects of type ostream are not copyable your code will not be valid in C++0x.

See here: http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#446

---

You can use the conditional operator this way instead:

#include <fstream>
#include <iostream>

int main(int argc, char* argv[])
{
    using namespace std;
    ofstream file("tmp.txt");
    int i=1;
    (i > 1 ? file : cout) << cin.rdbuf();
    return 0;
}

This compiles for me on gcc 4.4.3.

Not quite the one-liner you were looking for, I know.

NOTE: As Evan pointed out in his comment, this will result in the file being created even if not needed.

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I can see two possibilities here:

1. cout is of ostream type, the other (temporary) is ofstream, and they are not convertible to each other in any way
2. even if they are convertible the result is a temporary rvalue and it will not bind to operator<< as it requires nonconst lvalue reference.