Question:
I accidentally found:
cout << cout;
The output is some address. What does this address mean, and why is it shown?
I am looking this question.
Answer:
Because ostream
overload operator
, and that's the closes match for the call to
void*()operator
, the result of the cast
<<(void*)cout
is
printed. Which in your case is that address. Remember that cout
is
an object.
Basically the call translates to:
cout.operator<<((void*)cout);