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This might be a beginner question and understanding how cout works is probably key here. If somebody could link to a good explanation, it would be great.
cout<<cout and cout<<&cout print hex values separated by 4 on a linux x86 machine.
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cout in C++ It is used to display the output to the standard output device i.e. monitor. It is associated with the standard C output stream stdout. The data needed to be displayed on the screen is inserted in the standard output stream (cout) using the insertion operator(<<).
std::cout is used to output a value (cout = character output) std::cin is used to get an input value (cin = character input) << is used with std::cout, and shows the direction that data is moving (if std::cout represents the console, the output data is moving from the variable to the console).
cout is a reserved word. print is not, in c++. cout is the standard function to write something on the standard OUTput (screen, nowadays, printer before). Formely print, tells a programm to print on printer, cout on the screen.
cout << cout is equivalent to cout << cout.operator void *(). This is the idiom used before C++11 to determine if an iostream is in a failure state, and is implemented in std::ios_base; it usually returns the address of static_cast<std::ios_base *>(&cout).
cout << &cout prints out the address of cout.
Since std::ios_base is a virtual base class of cout, it may not necessarily be contiguous with cout. That is why it prints a different address.
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cout << cout is using the built-in conversion to void* that exists for boolean test purposes. For some uninteresting reason your implementation uses an address that is 4 bytes into the std::cout object. In C++11 this conversion was removed, and this should not compile.
cout << &cout is printing the address of the std::cout object.
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