Passing a function object to a constructor [duplicate] - c++

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Pass anonymous function object to std::function?
(1 answer)
Closed 7 years ago.
What I am trying to achieve is to make a functor that can take different functors as arguments.
Edit: the reason for my problem, the "most vexing parse", and the solution are well-described: see this question and answer, the whole most-vexing-parse tag, and even the wikipedia page. Still, I was not able to identify the problem before asking, and will leave this question as it might help others.
What I did:
In a header file functor.hpp:
#ifndef FUNCTOR_HPP
#define FUNCTOR_HPP
#include <functional>
template <typename T, typename BinOp = typename std::plus<T>>
struct doer {
BinOp op;
doer(BinOp o = std::plus<T>()) : op(o) {}
T operator()(const T& a, const T& b) const
{ return op(a, b); }
};
#endif // FUNCTOR_HPP
With this header, I can write a program functor.cpp like this:
#include <iostream>
#include "functor.hpp"
int main()
{
doer<int> f;
std::cout << f(3, 7) << std::endl;
}
and I can compile and run it to get, as expected:
$ make functor
g++ -std=c++14 -pedantic -Wall functor.cpp -o functor
$ ./functor
10
$
I am struggling to find a way to instantiate my doer with a different operator (not std::plus<T>).
doer<int, std::multiplies<int>> f2(std::multiplies<int>());
This compiles without a problem, but I have not been able to figure out a way to call f2(3, 7), to get the product 21. For example, if I add another line to the program:
int r = f2(3, 7);
and try to compile, I get:
$ make functor
g++ -std=c++14 -pedantic -Wall functor.cpp -o functor
functor.cpp: In function ‘int main()’:
functor.cpp:10:20: error: invalid conversion from ‘int’ to ‘std::multiplies<int> (*)()’ [-fpermissive]
int r = f2(3, 7);
^
functor.cpp:10:20: error: too many arguments to function ‘doer<int, std::multiplies<int> > f2(std::multiplies<int> (*)())’
functor.cpp:9:37: note: declared here
doer<int, std::multiplies<int>> f2(std::multiplies<int>());
^
functor.cpp:10:20: error: cannot convert ‘doer<int, std::multiplies<int> >’ to ‘int’ in initialization
int r = f2(3, 7);
^
What is going on? Seems almost like f2(3, 7) somehow is not calling the overloaded operator()...

Answer 1

Most vexing parse. Try this:
doer<int, std::multiplies<int>> f2((std::multiplies<int>()));
or this:
doer<int, std::multiplies<int>> f2 = std::multiplies<int>();
or this:
doer<int, std::multiplies<int>> f2{std::multiplies<int>()};