How do i write a pointer-to-member-function with std::function?

Question

I know how to declare int fn(double) inside of std::function ( std::function<int(double)> ). I know how to write a pointer-to-member-function ( typedef int (A::*MemFn)(double d); ). But how do i write a pointer-to-member-function with std::function?

Dummy code if you feel like compiling/testing

-edit- based on answers i think i'll just use the typedef and not bother with std::function

#include <cstdio>
#include <functional>

struct A{ int fn(double){ return 0; } };
int fn2(double){ return 0; }

typedef int (A::*MemFn)(double d);
typedef std::function<int(double)> MemFn2;

void Test(A*a, MemFn2 fn){
    fn(1.2f);
}
void Test(A*a, MemFn fn){
    (a->*fn)(1.2f);
}

int main(){
    Test(new A, &A::fn);
    Test(new A, &fn2);
}

Answer 1

std::function is perfectly capable of storing a member function pointer directly. However, you have to adjust the argument list appropriately. Member pointers must be called with an instance of the type (or a derived type). When putting them in a std::function , the first argument in the argument list is expected to be a pointer (or reference or smart-pointer) to the object type.

So, if I have the following class:

struct Type
{
public:
    int Foo();
};

The correct syntax to store this member function in a std::function is:

std::function<int(Type&)> fooCaller = &Type::Foo;

If you want to preserve the argument list (in your case, int(double) ), then you need to provide the instance outside of the function . This can be done via std::bind :

struct A{ int fn(double){ return 0; } };

A anInstance;
std::function<int(double)> fnCaller = std::bind(&A::fn, &anInstance, std::placeholders::_1);

Note that it is your responsibility to ensure that the object pointer you provide to std::bind remains alive so long as fnCaller is alive. If you return fnCaller to someone, and it has a pointer to a stack object, you're in trouble.

What's nice is that you could bind a shared_ptr (or any copyable smart pointer) as your object, thanks to how the function call mechanism is defined:

struct A{ int fn(double){ return 0; } };

auto anInstance = std::make_shared<A>();
std::function<int(double)> fnCaller = std::bind(&A::fn, anInstance, std::placeholders::_1);

Now you don't have to worry; the binder will continue to keep the object alive, since it stores a shared_ptr by value.

Answer 2

A member function is not a function. It is not itself anything you can call. All you can do is call a member function of an instance object . Only the pair of pointer-to-member-function and object constitutes a callable entity.

To bind an instance to a PTMF and obtain something callable, use bind :

#include <functional>

struct Foo
{
    double bar(bool, char);
};

Foo x;
using namespace std::placeholders;
std::function<double(bool, char)> f = std::bind(&Foo::bar, x, _1, _2);
f(true, 'a'); //...

As with lambdas, bind expressions have an unknowable type, and the conversion to std::function (as well as the actual dispatch) is potentially expensive. If possible, it is preferable to use auto for the type of the bind expression.

Answer 3

One of the guidelines in Scott Meyer's Modern C++11 book is to avoid std::bind and always use a lambda closure instead:

struct A{ int fn(double){ return 0; } };

std::function<int(double)> f = [a = A{}](double x) mutable { return a.fn(x); };

The mutable is necessary here, as the capture a might potentially be changed by the function call (since A::fn is non-const).

Answer 4

You can use std::binder1st to bind member function to a class instance:

typedef std::binder1st<std::mem_fun1_t<int, A, double>> MemFn;

void Test(A* a, double d)
{
   MemFn fn(std::mem_fun(&A::fn), a);
   int nRetVal = fn(d);
}

int main()
{
   Test(new A, 1.2f);
   return 0;
}

Answer 5

If you can use Boost then you can use Boost.Bind . It's easily accomplished like this:

boost::bind(&MyClass::MemberFunction, pInstance, _1, _2)

Hopefully it's fairly self-explanatory. _1 and _2 are placeholders for parameters you can pass through to the function.