"<<" ?

[James Parsons]

in cout what exactly is "<<"

 #include <iostream>
 #include <cstdio>

 int main(int argc, const char *argv)
 {
  std::cout<<"Hello, World!"<<std::endl;
  getchar();
 
  return 0;
 }

[David Parker]

if your going to do C++ then you shoud know what "<<" is

[Fransje]

if your going to do C++ then you shoud know what "<<" is

That's one of the most useful answers I've ever heard. Congrats, David!

'<<' is the bitwise left shift operator: http://www.learncpp.com/cpp-tutorial/38-bitwise-operators/
But in the case of std::cout it's overloaded: http://www.cplusplus.com/reference/ostream/ostream/operator%3C%3C/

[amchinese]

For "cout", "<<" means output. For numbers, "<<" means Shift operations. For class, you can define it yourself.("cout" is a class witch define in the file "iostream.h")

[MiiNiPaa]

<< is a bitwise left shift operator. It is overloaded to work differently with ostream and derived classes. Standard library provides overloads fo all built-in types and several calsses from standard library (std::string for example). You can also provide overload for your own classes.
Typical overload looks like:

std::ostream& operator<<(std::ostream& lhs, T const& rhs)
{
    //Do output here
    return lhs;
}

Where T is your type.
You need to return stream to allow for operator chaining. Suppose you have following code:

std::cout << "x" << 2;

As operator << has left-to-right associativity, it could be imagined as:

operator<<( operator<<(std::cout, "x"), 2);

Result of inner operator<< call is std::cout, so outer one will work fine.