How pass std::endl to a function and use it?

Question

I want to figure out how to pass a manipulator like std::endl to a function and then use the passed-in manipulator in the function. I can declare the function like this:

void f(std::ostream&(*pManip)(std::ostream&));

and I can call it like this:

f(std::endl);

That's all fine. My problem is figuring out how to use the manipulator inside f. This doesn't work:

void f(std::ostream&(*pManip)(std::ostream&))
{
  std::cout << (*pManip)(std::cout);            // error
}

Regardless of compiler, the error message boils down to the compiler not being able to figure out which operator<< to call. What do I need to fix inside f to get my code to compile?

Egocentrism

possible duplicate #1134888 – Fons

I don't think it's the same question. I don't want to create a custom manipulator, I just want to use an existing manipulator that's passed as an argument to a function. – Egocentrism

just call (*pManip)(std::cout) inside – Fons

Just go std::cout << pManip; – Lentissimo

std::cout << pManip will print the address of the function. (pManip is a function pointer.) See Sveltely's answer for the proper code. – Egocentrism

function pointers can be used this way too.So std::cout << pManip; should be ok. Indeed on your own code the problem was that you were passing result of (*pManip)(std::cout) to <<(stream output function) – Fons

My mistake, you're right. Thanks for the education. – Egocentrism

We all do such errs.It is cause of not being attentive. Being not attentive can lead such minor errors.Plus with c++ one should be a bit more attentive – Fons

Answer 1

void f(std::ostream&(*pManip)(std::ostream&))
{
  std::cout << "before endl" << (*pManip) << "after endl"; 
}

or

void f(std::ostream&(*pManip)(std::ostream&))
{
  std::cout << "before endl";
  (*pManip)(std::cout);
  std::cout << "after endl"; 
}

Bryannabryansk

Note that a const function pointer is also permitted: e.g. void f(std::ostream&(* const pManip)(std::ostream&)). – Carburet