I am new to C++ and I am confused about std::endl. When trying to understand what std::endl is, I had faced some resources which told me that it is a function.
However, how can a function be deprived of parentheses?
However, how can a function be deprived of parentheses?
The name of a function, without being followed by (), is just a reference to that function. It's exactly the same as with any other type:
void foo(int) {}
char x = 'a';
char *p = &x;
int main()
{
p; // Refers to p
*p; // Dereferences p (refers to whatever p points to)
foo; // Refers to foo
foo(42); // Calls foo
}
std::endl is a function (actually a function template) which takes one parameter of type "a stream", and works by inserting an EOL representation into that stream and then flushing it. You can actually use it like any other function, if you want to:
std::endl(std::cout);
The final piece of the puzzle is that the standard library provides an overload (again, a template) of operator << such that the LHS argument is a stream and the RHS argument is a function; the implementation of this operator calls the RHS argument (the function) and passes it the LHS one (the stream). Conceptually, there's something like this:
Stream& operator<< (Stream &s, const Function &f)
{
f(s);
return s;
}
Therefore, calling std::cout << std::endl invokes that operator overload, which in turn invokes std::endl(std::cout), which does the EOL insertion + flushing.
As to which form is to be preferred (direct call vs. << operator), it's definitely the use of <<. It's idiomatic, and it allows easy composition of multiple stream manipulators within a single expression. Like this:
std::cout << "Temperature: " << std::fixed << std::setprecision(3) << temperature << " (rounds to " << std::setprecision(1) << temperature << ')' << std::endl;