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?

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. — KnowItAllWannabe
– KnowItAllWannabe, Commented Apr 8, 2014 at 5:21
std::cout << pManip will print the address of the function. (pManip is a function pointer.) See Sveltely's answer for the proper code. — KnowItAllWannabe
– KnowItAllWannabe, Commented Apr 8, 2014 at 5:28

Sveltely · Accepted Answer · 2014-04-08 05:23:59Z

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"; }

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

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