模糊的C ++运算符重载

Question

I have the following code:

#include <iostream>

using namespace std;

ostream& f(ostream& os) {
    return os << "hi";
}

int main() {
    cout << "hello " << f << endl;
    return 0;
}

And somehow this works - the output is "hello hi". How does this get interpreted by the compiler? I don't understand how a function can be inserted into a stream.

Answer 1

std::ostream has an operator<< overload that receives a pointer to a function with the signature such as the one you wrote ( number 11 in this list ):

basic_ostream& operator<<(
    std::basic_ostream<CharT,Traits>& (*func)(std::basic_ostream<CharT,Traits>&) );

which just calls the given function passing itself as argument. This overload (along with several similar others) is there to allow you to implement stream manipulators , ie stuff that you output in the stream with an << and changes the state of the stream from there. For example, our version of the (incorrectly) ubiquitous std::endl may be implemented as

std::ostream &myendl(std::ostream &s) {
    s<<'\n';
    s.flush();
    return s;
}

which can then be used exactly as the "regular" std::endl :

std::cout<<"Hello, World!"<<myendl;

(the actual implementation is templated and a bit more complicated because it has to work even with wide streams, but you got the idea)

Answer 2

std::ostream::operator<< has an overload which accepts a function as parameter; and the body of that overload is to invoke the function given.

This is exactly how endl works in fact. endl is actually a function similar to:

ostream &endl(ostream &os)
{
    os << '\n'; 
    os.flush(); 
    return os;
}