All,
Why does the following code fail to compile for 'std::endl', but it's fine for all of the other inserted types?
#include <sstream> // ostringstream
/// @brief A class that does streamed, formatted output via 'operator<<'.
class My_Stream
{
public:
/// @brief A member method that manipulates the underlying stream.
void foo()
{
m_oss << "foo_was_here; ";
}
private:
/// @brief The underlying stream.
std::ostringstream m_oss;
/// @brief 'operator<<' is a friend.
template< typename T >
friend My_Stream& operator<<( My_Stream& a_r_my_stream,
const T& a_r_value );
};
/// @brief A manipulator that calls a class method.
My_Stream& manipulator_foo( My_Stream& a_r_my_stream )
{
a_r_my_stream.foo();
return a_r_my_stream;
}
/// @brief The generic insertion operator.
template< typename T >
My_Stream& operator<<( My_Stream& a_r_my_stream,
const T& a_r_value )
{
a_r_my_stream.m_oss << a_r_value;
return a_r_my_stream;
}
/// @brief Define an iostream-like manipulator for my-stream.
typedef My_Stream& ( * my_stream_manipulator ) ( My_Stream& );
/// @brief The specialized 'my_stream_manipulator' insertion operator.
template<>
My_Stream& operator<<( My_Stream& a_r_my_stream,
const my_stream_manipulator& a_r_manipulator )
{
return a_r_manipulator( a_r_my_stream );
}
int main( int argc, char* argv[] )
{
My_Stream my_stream;
my_stream << 'c'; // char
my_stream << "string"; // c-string
my_stream << 1u; // unsigned int
my_stream << -1; // signed int
my_stream << 5.3f; // float
my_stream << -23.345; // double
my_stream << std::boolalpha; // std::ios_base manipulator
my_stream << std::endl; // std::ostream manipulator
my_stream << manipulator_foo; // my_stream manipulator
return 0;
}
I get the following G++ 4.5 error:
willo:~/test_cpp$ g++ -Wall test_overloaded_insertion_manipulators.cpp test_overloaded_insertion_manipulators.cpp: In function ‘int main(int, char**)’: test_overloaded_insertion_manipulators.cpp:60: error: no match for ‘operator<<’ in ‘my_stream << std::endl’
I expect the code to instantiate a 'operator<<' for std::endl, just like it did for the primitives, std::ios_base and my custom manipulator.
For context, I'm trying to create a light-API IOStream-like class that works with current IOStream manipulators, as well as one or two more custom manipulators.
Because endl
is a function template:
template <class charT, class traits>
basic_ostream<charT,traits>& endl(basic_ostream<charT,traits>& os);
So the identifier itself is not a value. It only becomes a value (function pointer) when it's instantiated. But your operator<<
is itself a template, and there is no type information available to the compiler to decide which types to instantiate endl
with.
In contrast, e.g. boolalpha
is:
ios_base& boolalpha(ios_base& str);
Hence why it works.
endl
works for basic_ostream
, because that one defines operator<<
overloads as member functions taking function pointers; in particular:
basic_ostream<charT,traits>& operator<<(basic_ostream<charT,traits>& (*pf)(basic_ostream<charT,traits>&));
So in a call like stream << endl
, it would know charT
and traits
from type of this
(i.e. left side of operator), and that would give it exact type of function pointer to expect on the right side - which it would then use to instantiate the corresponding version of endl
. You can do the same for your class.
You need to define stream manipulators as a separate friend.
Add this friend:
// Note: Untested (don't have a compiler here).
template <class charT, class Traits>
friend My_Stream& operator<<( My_Stream&, std::basic_ostream<charT, Traits>& (*)(std::basic_ostream<charT, Traits>&));
Then you need to define the function:
template <class charT, class Traits>
My_Stream& operator<<( My_Stream& stream, std::basic_ostream<charT, Traits>& (*manip)(std::basic_ostream<charT, Traits>&))
{
(*manip)(stream.m_oss);
return stream;
}
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