Operator << in a user defined method - c++

Question

I was trying to write a method that would take parameters using the >> operator as parameters similar to std::cin but I don't know how. Is it possible to create this kind of method that would take this stream like parameter, convert it properly (for example convert all ints to strings etc) and then save to an std::string variable?
Here is an example of how I would like to run the function:
int i = 0;
myMethod << "some text" << i << "moar text";
Inside that method I would like to take those parameters and store in a string.
Edit
I will try to explain exactly what this application is about: I Am trying to make a Clogger singleton class which will be used to save logs to a file. With this construction, I can call *CLogger::instance() << "log stuff"; from anywhere in the code and that's OK. Thanks to answers from this topic I have come to this. The problem is that each operator<< I use, then the object is going to be called. So if I do *CLogger::instance() << "log stuff " << " more stuff " << " even more";` this method(?) is going to be called 3 times:
template<typename T>
CLogger& operator<<(const T& t)
{
...
return *this;
}
That's not good for me as I intend to add some text before and after each log line. For example I would always like to add time before and std::endl after. Following the example I gave instead of getting:
[00:00] log stuff more stuff even more
I would get:
[00:00] log stuff
[00:00] more stuff
[00:00] even more
So I made an attempt to remove this behaviour by changing the method like this:
template<typename T>
CLogger& operator<<(const T& t)
{
ostringstream stream;
stream << t;
m_catString += stream.str();
if (stream.str() == "\n")
{
push_back(m_catString);
m_catString.clear();
}
return *this;
}
This way the program knows when to push new log line if I add "\n" at the end. Its nearly ok, as I bet I will forget to add this. Is there any more clever way?

Answer 1

You can't pass parameters to a method using <<, you need an object.
Something like this:
struct A
{
template<typename T>
A& operator<<(const T& t)
{
std::ostringstream stream;
stream << t;
data += stream.str();
return *this;
}
std::string data;
};
// ...
A a;
a << "Hello " << 34 << " World";
std::cout << a.data;
Regarding your update:
The most obvious thing is to implement the operator in CLogger instead and get rid of the Pusher class; that would let you write *CLogger::instance() << "sample text" << 10;
If you can't do that for some reason (you're giving out information piecemeal, so it's hard to tell), you can declare Pusher a friend and use the same method as everywhere else:
struct Pusher
{
template<typename T>
Pusher& operator<<(const T& t)
{
std::ostringstream stream;
stream << t;
CLogger::instance()->push_back(stream.str());
return *this;
}
};

Answer 2

The only way I know is creating a class class Method and then overload the operator<<, operators overloading
template<class T>
Method &operator<<(const T &x)
{
// Do whatever you like
return *this;
}
Then you can use it like :
Method myMethod;
myMethod << ... ;
you can look at this question about creating a cout-like class
std::cin and std::cout are not functions by the way
EDIT
class CLogger
{
...
template<typename T>
CLogger& operator<<(const T& t)
{
push_back(std::to_string(t));
return *this;
}
};
You don't have to create a class Pusher, just overload the operator in your first class, now you can use it with your object :
myCLogger << t; // this would call the function push back