Why insertion operator is printing address instead of string?

Question

I have simple lines of code, where I am using insertion operator << to show hello world string. If I use a operator b then it should result to a.operator(b); I try to do same thing with insertion operator and in output I got address of string, rather than actual string.

std::cout<<"Hello world"<<std::endl;
std::cout.operator<<("Hello world").operator<<(std::endl);

Output:

Hello world
0120CC74

I am using Visual Studio.

Does my operator conversion has any problem?

Answer 1

std::cout<<"Hello world"<<std::endl;

use overloaded output operator for const char*, that is free function, not member function.

std::cout.operator<<("Hello world").operator<<(std::endl);

use overloaded output operator for const void*, since const char* is implicitly convertible to const void*.

You can look at member overloads here and free overloads here

Answer 2

My bet is that member function operator for std::ostream(char*) is not overloaded.

If you look at ostream::operator<<, void* is best match and char* naturally gets converted to it, while global operator<<(std::basic_ostream), has exact overloads for char* types, which gets picked up.

Of course, they behave differently.