Returning constructor arguments in braces?

Question

The brace initialization (or uniform init) really is confusing. What exactly happens when a function returns a list of arguments in braces?

std::vector<double> foo()
{
    //is this the same as: std::vector<double>{1, 2}
    //or std::vector<double>(1, 2)?
    //or something else?
    return {1, 2};
}

Answer 1

return {1, 2};, the return value is list-initialized from {1, 2}, as the effect, the returned std::vector<double> contains 2 elements with value 1 and 2.

return std::vector<double>{1, 2};, the return value is copy-initialized from std::vector<double>{1, 2}, as the effect, the returned std::vector<double> contains 2 elements with value 1 and 2. In concept it'll construct a temporary std::vector<double> and the return value is copy-initialized from the temporary; because of mandatory copy elision (since C++17) the copy/move operation is ommitted and the effect is exactly the same as the 1st case.

return std::vector<double>(1, 2), the return value is copy-initialized from std::vector<double>(1, 2), as the effect, the returned std::vector<double> contains 1 elements with value 2. Mandatory copy elision takes effect in this case too.