问题
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.
回答1:
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.
回答2:
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.
回答3:
cout << &cout is passing cout the address of cout.
cout << cout is printing the value of implicitly casting cout to a void* pointer using its operator void*.
回答4:
As already stated, cout << cout uses the void* conversion provided for bool testing (while (some_stream){ ... }, etc.)
It prints the value &cout + 4 because the conversion is done in the base implementation, and casts to its own type, this is from libstdc++:
operator void*() const
{ return this->fail() ? 0 : const_cast<basic_ios*>(this); }
回答5:
cout<<&cout is passing the address of cout to the stream.
来源：https://stackoverflow.com/questions/7489069/whats-the-difference-between-coutcout-and-coutcout-in-c