Typo with “cout < myint”. Why does it work?

Question

I have this code and I searched for hours why it fails to print my income

int const income = 0;
std::cout << \"I\'m sorry, your income is: \" < inco         


        

Answer 1

integral constant 0 is also a null pointer constant - it can be compared to the result of ostream's operator void *. Note that it'll fail if the constant has any value but 0.