CODEWITHSUNDEEP
c++stl
Kill_Switch
Kill_Switch

Reputation: 93

C++ std::list with std::function: calling remove() won't build

can someone explain me why this code won't build:

std::function<void(int)> intcb = [](int a)
{
    std::cout << "callback: " << a << std::endl;
};

std::list<std::function<void(int)>> list;
list.push_back(intcb);
list.remove(intcb);

build error located in remove function:

no match for 'operator==' (operand types are 'std::function<void(int)>' and 'const value_type' {aka 'const std::function<void(int)>'})

thanks for your help

Upvotes: 3

Views: 441

Answers (2)

Uladzislau Shablinski
Uladzislau Shablinski

Reputation: 174

Does this make sense: what you have in your list is function object and while you put it to the list you're actually copying the intcb. So you need a way to compare the copies. Instead you can try storing by pointer, so removing will work by comparing addresses which works out of the box.

#include <iostream>
#include <list>
#include <functional>

std::function<void(int)> intcb = [](int a)
{
    std::cout << "callback: " << a << std::endl;
};


int main() {
    std::list<std::function<void(int)>*> list;
    list.push_back(&intcb);
    list.remove(&intcb);
    return 0;
}

Upvotes: 0

Nikita Kovalev
Nikita Kovalev

Reputation: 51

Well, you want to create a list of functions and then enable "remove" functionality, but for removal operation list needs to be able to compare instances of its type, because it does not rely on references to objects.

So you need to somehow define operator== for your specified function type

Upvotes: 5

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