what does compiler when operator<<(std::basic_ostream) overloaded as friend [duplicate]

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• What are the basic rules and idioms for operator overloading? 7 answers

declaration in class:

friend ostream& operator<<(ostream & os, const MyClass & _my_class );

call:

MyClass mc;
...
...
cout << mc << endl;

1. Why does something like that not work?

   cout.operator<<(mc)

2. Where is the first argument?

   (ostream &os, ...)

3. How does compiler solve this?