Using erase-remove idiom for function<void()> - c++

Question

Stacked people.
Iam trying to implement an observer(esque?) pattern for my program. I have a component which stores what functions should be called if an event occours. My prolem is that i dont know how should i erase my function from the container, if the need arises. Tried storing the functions by reference, but iam not sure how to do that(or if thats possible.)
#include <map>
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
enum class EVENT_TYPE{
anEvent
};
class EventableComponent{
map<EVENT_TYPE, vector<function<void()>>> listeners;
public:
void trigger(EVENT_TYPE _et){
for(auto& it : listeners[_et]){
it();
}
}
void registerListener(EVENT_TYPE _et, function<void()> _fn){
listeners[_et].push_back(_fn);
};
void removeListener(EVENT_TYPE _et, function<void()> _fn){
//error C2678: binary '==' : no operator found which takes a left-hand operand of type 'std::function<void (void)>'
//(or there is no acceptable conversion)
listeners[_et].erase(remove(listeners[_et].begin(), listeners[_et].end(), _fn), listeners[_et].end());
};
};
int main(){
EventableComponent ec;
// this would become a member function for a class somewhere down the line
auto fn = [](){cout << "Hello.\n"; };
ec.registerListener(EVENT_TYPE::anEvent, fn);
ec.trigger(EVENT_TYPE::anEvent);
ec.removeListener(EVENT_TYPE::anEvent, fn);
ec.trigger(EVENT_TYPE::anEvent);
cin.get();
return 0;
};

Answer 1

Your problem can be reduced to the fact that two std::function instances cannot be compared for equality. std::remove requires operator==, and std::function does not have it. See "Why is std::function not equality comparable?".
Consider the following situation.
Let's say you defined two lambdas in your main:
auto fn = [](){cout << "Hello.\n"; };
auto fn2 = [](){cout << "Hello.\n"; };
Now, are those two equal or not? They do the same thing, but perhaps this is sheer coincidence. Would they become unequal if the second "Hello" became "Hello2"? Would they become unequal if the second one was no longer a lambda but a real function void f()?
The thing is that there can be no generally useful definition of equality for function objects, so it's up to you to define what equality really means in the context of your program.
You have several options to solve the problem at hand. One would be to operate on pointers to std::function objects. Pointers can be compared, and proper use of std::unique_ptr makes sure that deallocation is handled correctly.
Or you assign an identifier to every std::function you use. See the following modified example of your code in which direct storage of std::function<void()> in the vector is replaced with a custom type EventFunction that maps an int to the function object. The example uses std::remove_if to compare only the ints:
#include <map>
#include <vector>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
enum class EVENT_TYPE{
anEvent
};
struct EventFunction {
function<void()> f;
int id;
};
class EventableComponent{
map<EVENT_TYPE, vector<EventFunction>> listeners;
public:
void trigger(EVENT_TYPE _et){
for(auto& it : listeners[_et]){
it.f();
}
}
void registerListener(EVENT_TYPE _et, EventFunction _fn){
listeners[_et].push_back(_fn);
};
void removeListener(EVENT_TYPE _et, int function_id){
//error C2678: binary '==' : no operator found which takes a left-hand operand of type 'std::function<void (void)>'
//(or there is no acceptable conversion)
listeners[_et].erase(remove_if(listeners[_et].begin(), listeners[_et].end(),
[&](EventFunction const& e) { return e.id == function_id; }), listeners[_et].end());
};
};
int main(){
EventableComponent ec;
// this would become a member function for a class somewhere down the line
auto fn = [](){cout << "Hello.\n"; };
ec.registerListener(EVENT_TYPE::anEvent, EventFunction{ fn, 1 });
ec.trigger(EVENT_TYPE::anEvent);
ec.removeListener(EVENT_TYPE::anEvent, 1);
ec.trigger(EVENT_TYPE::anEvent);
};
Tried storing the functions by reference, but iam not sure how to do
that(or if thats possible.)
It's not possible because you cannot store references in standard-library containers. But I suppose the idea is similar to the one with pointers I mentioned above.