c++ - Why insertion operator is printing address instead of string? -

i have simple lines of code, using insertion operator << show hello world string. if use a operator b should result a.operator(b); try same thing insertion operator , in output got address of string, rather actual string.

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

output:

hello world
0120cc74

i using visual studio.

does operator conversion has problem?

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

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

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

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

you can @ member overloads here , free overloads here