如何比较 std::function 对象？

Question

I have a vector of std::function objects defined like this:

std::vector<std::function<void()>> funcs = { myFunc1, myFunc2, myFunc3 };
//The functions are defined like this:
//void myFunc1() { ... }

I am trying to search through this array for a specific function. My first attempt was using the std::find function, like this:

auto iter = std::find(funcs.begin(), funcs.end(), myFunc2);
//CS2679: binary '==': no operator found which takes a right-hand operator of type '_Ty (__cdecl &)' or there is no acceptable conversion

I learned the hard way that std::function::operator==() does not compare equal unless the function objects are empty (clearly not the case). So I tried using std::find_if to make use of the std::function::target() method:

auto iter = std::find_if(funcs.begin(), funcs.end(), [](const std::function<void()>& f)
    {
        if (*f.target<void()>() == myFunc2)
        //using or not using that '*' before f in the condition makes no difference in the error
            return true;
        return false;
    });

My compiler (VC++ 2019) still complained with the same exact error. Intrigued, I tried writing a find function by hand to see what is happening, but I had no success, got the same error:

auto iter = funcs.begin();
for (; iter != funcs.end(); iter++)
    if (*iter->target<void()>() == myFunc2)
        break;

So here is the question. How can I compare 2 std::function objects to see if they store the same function?

Answer 1

As shown here , template member function target accepts type that is compared with a stored object type. In you case it is a pointer to a function. You have to change

*iter->target<void()>() == myFunc2

to

*iter->target<void(*)()>() == myFunc2

Note that this will only only you to find a plain C function, not an arbitrary callable object (like a lambda function). I think you should consider using plain pointers instead of std::function here.