Re: Implementing operator<<
On 17 Led, 15:34, soy.h...@gmail.com wrote:
Hi all
I have a class StreamLogger which implements operator << in this way:
template <class T> StreamLogger& operator<<(const T& t)
{
<print the stuff using fstream, etc.>
return *this;
}
It works perfectly, using it like:
m_logger << "My ID is " << getID() << "\n";
but for some reason it does not compile this code:
m_logger << std::endl;
It gives the following error:
--------------------------------------------------------------------------------
.\src\Dealer.cpp(70) : error C2678: binary '<<' : no operator found
which takes a left-hand operand of type 'StreamLogger' (or there is no
acceptable conversion)
C:\Program Files\Microsoft Visual Studio 8\VC\atlmfc\include
\afx.h(1428): could be 'CDumpContext &operator <<(CDumpContext
&,ATL::CTimeSpan)'
...
--------------------------------------------------------------------------------
Could you guys help me understand why it doesn't compile? (because I
dont have a clue...)
Thanks!!!
Jorge
You have to do following steps:
1. Create your own manipulator in std:
namespace std
{
StreamLogger& endl(StreamLogger& s)
{
// Here do what manipulator should do
return s;
}
StreamLogger& hex(StreamLogger& s)
{
// Here do what manipulator should do
return s;
}
// Define all standard manipulators which do not have parameters
// ...
}
2. Overload operator<< to accept function of the same type as
manipulator is:
StreamLogger& operator<<(StreamLogger& s, StreamLogger& (*manip)
(StreamLogger& s))
{
return manip(s);
}
3. Use your manipulator in usual way:
StreamLogger sl;
sl << std::endl;
You should define all standard manipulators for your StreamLogger to
have standard interface. Do not ask me how to support anipulators with
parameters. I know it is (of course) possible, but I do not remember
the solution.