How is the return type of a ternary operator determined?

Question

I was solving a problem about movement of a bishop on a chessboard. At one point of my code, I had the following statement:

std::cout << (abs(c2-c1) == abs(r2-r1)) ? 1 : 2 << std::endl;

This generates the following error:

error: invalid operands of types 'int' and '<unresolved overloaded function type>' to binary 'operator<<'

However, I instantaneously fixed this error by including an additional variable in my code:

int steps = (abs(c2-c1) == abs(r2-r1)) ? 1 : 2;
std::cout << steps << std::endl;

How does the ternary operator work, and how is its return type determined (as the compiler called it <unresolved overloaded function type>)?

Answer 1

This has nothing to do with how the return type is deduced and everything to do with operator precedence. When you have

std::cout << (abs(c2-c1) == abs(r2-r1)) ? 1 : 2 << std::endl;

it isn't

std::cout << ((abs(c2-c1) == abs(r2-r1)) ? 1 : 2) << std::endl;

because ?: has lower precedence than <<. That means what you actually have is

(std::cout << (abs(c2-c1) == abs(r2-r1))) ? 1 : (2 << std::endl);

and this is why you get an error about an <unresolved overloaded function type>. Just use parentheses like

std::cout << ((abs(c2-c1) == abs(r2-r1)) ? 1 : 2) << std::endl;

and you'll be okay.