Comparing std::function for member functions

Question

I tried to search and here similar questions:
question 1
question 2

But anyway, I can't compare member functions. Here's an example:

class ClassA { public: int add(int a, int b) { return a + b; } }; int main() { ClassA a1{}; function<int(int, int)> f1 = bind(&ClassA::add, a1, placeholders::_1, placeholders::_2); function<int(int, int)> f2 = bind(&ClassA::add, a1, placeholders::_1, placeholders::_2); cout << boolalpha << "f1 == f2 " << (f1.target_type() == f2.target_type()) << endl; // true cout << (f1.target<int(ClassA::*)(int, int)>() == nullptr) << endl; // true return 0; }

From the code it's obvious that f1 and f2 are different. The first cout shows true because the types are the same, it's ok. But why the second cout is true? Why function::target() returns nullptr?

P.S.: I want to create a simple delegate system so I can pass any function (global, static, member) around. With std::function I can add a callback, but I don't know how to remove it.

Barry · Accepted Answer · 2015-01-06 12:08:07Z

That's because f1's target type is not int(ClassA::*)(int, int). Its target type is going to be the result of that bind expression, which on gcc happens to be:

std::_Bind<std::_Mem_fn<int (ClassA::*)(int, int)> ( ClassA, std::_Placeholder<1>, std::_Placeholder<2>)>

Which you can see using the ABI demangler:

#include <cxxabi.h> // note, the following line technically leaks, but // for illustrative purposes only it's fine cout << abi::__cxa_demangle(f1.target_type().name(), 0, 0, 0) << endl;

Note that if the target type was actually a class method, you wouldn't be able to call it with two ints - you'd also need the ClassA*. For instance, this function's target type is int(ClassA::*)(int, int):

function<int(ClassA*, int, int)> f3 = &ClassA::add;

int(ClassA*, int, int) is not itself a member function type. So the std::function target type is not a member function type in that case. This affects the result of the target member function.
@Cheersandhth.-Alf f3.target_type() == typeid(&ClassA::add)
you're right, sorry. what was i thinking. +1 on your comment. :)

Yakk - Adam Nevraumont · Accepted Answer · 2015-01-06 12:04:16Z

Those std::functions do not hold member functions. They hold bind result types. As the bind is the same pattern of types, the bind result types are the same.

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