std::cout doen't like std::endl and string in conditional-if

Question

main.cpp: In function ‘void PrintVector(std::vector<std::__cxx11::basic_string<char> >&, bool)’:
main.cpp:16:41: error: overloaded function with no contextual type information
  std::cout << ((newline)? (std::endl) : "");
                                         ^~

Why std::cout doen't like std::endl and string in conditional-if?

Answer 1

std::endl is a stream manipulator. It's a function. It does not have a common type with "". So they cannot be the two types of a conditional expression. Since the common type is the type of the whole expression.

You probably don't even need everything std::endl does besides adding a new line, so just replace it with "\n" to print a new-line. That way the common type may be deduced to const char* after all the usual conversions are performed on the operands.

Answer 2

I changed it to:

std::cout << (newline? "\n" : "") << std::flush;

It is not possible to write it with ' (would be faster):

std::cout << (newline? '\n' : '') << std::flush;

because '' is empty and leads to "error: empty character constant".

The solution with the conditional-if is so complex that one should prefer the following:

if (newline) std::cout << std::endl;