This code gives me error in VS2015 update 1:
error C2893: Failed to specialize function template 'unknown-type
std::invoke(_Callable &&,_Types &&...)'
#include <iostream>
#include <functional>
using std::cout;
class A
{
public:
virtual void init()
{
cout << "A";
};
};
class B
{
public:
virtual void init()
{
cout << "B";
};
};
class C : private A, private B
{
std::function<void()> a_init = &A::init;
std::function<void()> b_init = &B::init;
public:
void call()
{
a_init();
b_init();
}
};
int main()
{
C c;
c.call();
return 0;
}
Any ideas if that's VS compiler is buggy or my code?
EDIT
#include "stdafx.h"
#include <functional>
class A
{
public:
virtual void inita()
{
cout << "A";
};
};
class B
{
public:
virtual void initb()
{
cout << "B";
};
};
class C : private virtual A, private virtual B
{
/*std::function<void()> a_init = &A::init;
std::function<void()> b_init = &B::init;*/
public:
void call()
{
inita();
}
};
You're trying to assign non-static member functions into a std::function taking no arguments. That cannot work, since non-static member functions have an implicit this parameter.
How to solve this depends on what you want to do. If you want to call the stored function on an arbitrary object supplied at call time, you'll need to change the std::function signature:
std::function<void(A*)> a_init = &A::init;
void call()
{
a_init(this); // or some other object of type A on which you want to invoke it
}
[Live example]
If, on the other hand, you want to call it without arguments, you will have to bind an object of type A into the std::function at initialisation:
std::function<void()> a_init = std::bind(&A::init, this);
void call()
{
a_init()
};
[Live example]
Change the function from virtual to static and the code will work. You need a specific instance of a class to call a non-static function.
On the other hand, if you wish to use non-static function, you can add the following constructor:
C(A &a, B &b)
{
a_init = std::bind(&A::init, &a);
b_init = std::bind(&B::init, &b);
}
and then use it in main like this:
A a;
B b;
C c(a, b);
c.call();
EDIT:
If public inheritance is acceptable option, then you can do it even simpler.
Constructor:
C()
{
a_init = std::bind(&A::init, this);
b_init = std::bind(&B::init, this);
}
Usage:
C c;
c.call();
Related
I have a nested class B inside class A. I need to pass a custom function with one integer argument and return type of void at runtime to class B.
This is how I tried to do it. First I passed the function through the constructor of A. And then to pass it to B, I tried to use pointer to member function. However I can't figure out how to call function foo() inside doStuff().
class A {
void(*f)(int);
A(void(*f)(int)) : f(f) {};
class B {
void(*A::*foo)(int) = &A::f;
void doStuff() {
var = 10;
*foo(var); //Doesn't work
}
};
};
void testFunction(int a) {
//do something
}
A a(testFunction);
What is the correct way to call it? And does this solution make sense as the whole?
Edit: Possible alternative solution:
class A {
A(void(*f)(int)) {
b = B(f);
}
class B {
void(*f)(int);
B() {}
B(void(*f)(int)) : f(f) {}
void doStuff() {
var = 10;
f(10);
}
};
B b;
};
Note that in order to call a pointer-to-function member of A you need an object of type A to call it on. In the example below, the reference to an A object was passed as an argument to B::doStuff.
#include <iostream>
class A
{
public:
void (*f)(int);
A(void (*f)(int)) : f(f) { }
class B {
public:
void (*A::*foo)(int) = &A::f;
void doStuff(A &a) {
(a.*foo)(10);
}
};
};
void testFunction(int a) {
std::cout << "inside testFunction(" << a << ")" << std::endl;
}
int main()
{
A a(testFunction);
A::B b;
b.doStuff(a);
}
Example output: inside testFunction(10).
Is it possible to pass this by default ?
Here is what I currently have
class A
{
public:
template<typename T>
void dowithT(T t) {}
};
class B
{
public:
A a;
B()
{
//Calling 'dowithT' with 'this'
a.dowithT(this);
}
};
This function requires passing this from the caller of the function every time. So I wondered if there is a way to encapsulate this task, so that you don't need to pass this to dowithT.
I tried to do something like this:
class A
{
public:
// '= this' doesn't compile
template<typename T>
void dowithT(T t = this) {}
};
class B
{
public:
A a;
B()
{
//Calling 'dowithT' without 'this'
a.dowithT();
}
};
Unfortunately, I can't use templates, so my first solution isn't an option.
Is this possible?
Edit: I gave a concrete answer with my own implementation below. Also with a few mor deatils of what I wanted in the end.
TL;DR No, this is not possible.
this is not the same type in every class, you can't generalize it, so no, not possible.
Additionally, what would this be if doWithT() was called from a non-member function? nullptr?
That's why it isn't possible. You have to use a template.
Instead of B having a member of type A, it can inherit from A, and use something like the "curiously recurring template pattern."
If you cannot make class A a template, you can still do it like so:
class A
{
protected:
template <class T>
void dowithT()
{
T* callerthis = static_cast<T*>(this);
// callerthis is the "this" pointer for the inheriting object
cout << "Foo";
}
};
class B : public A
{
public:
B()
{
dowithT<B>();
// Or A::dowithT<B>();
}
};
dowithT() must only be called by an inheriting class (hence I made it protected), with the template parameter the caller's own type, or you'll break everything.
You may achieve exactly what you want by using a private mixin class to provide the dowithT method that takes no arguments:
#include <iostream>
#include <typeinfo>
class A
{
public:
template<typename T>
void dowithT(T* t) {
std::cout << "Hello, World" << typeid(*t).name() << std::endl;
}
};
template<class Owner>
struct calls_a
{
void dowithT()
{
auto p = static_cast<Owner*>(this);
p->a.dowithT(p);
}
};
class B
: private calls_a<B>
{
friend calls_a<B>;
A a;
public:
B()
{
//Calling 'dowithT' with 'this'
dowithT();
}
};
int main()
{
B b;
}
No, it is not possible. There is nothing really special about this when used as an argument to a function taking T* (template or not), it's just a pointer like any other.
this A is different from this B. In your first code, this refers to the caller, while in the second this refers to the callee. Thus what you want to do isnt really possible.
Here's one possibility, which might, or might not suit your needs:
template<typename T>
class A
{
public:
A(T t) : t(t) {}
void dowithT()
{
cout << "Foo";
}
private:
T t;
};
class B
{
public:
A<B*> a;
B() : a(this)
{
a.dowithT();
}
};
You could use a private method in class B that acts as a relay, and use the constant nullptr as a special value for this, if you want to be able to pass other values:
class B
{
public:
A a;
B()
{
//Calling 'dowithT' with 'this'
innerdo();
}
private:
void innerdo(B *p = nullptr) {
if (p == nullptr) p = this;
a.dowithT(p);
}
};
If you only need to pass this it is even simpler
void innerdo() {
a.dowithT(this);
}
After trying out various things you mentioned, I'd like to give my answer/solution to the problem myself to clarify some details:
#include <iostream>
using namespace std;
#include <functional>
template <typename CallerType>
class AFunctionConstructor{
private:
virtual void abstr()
{}
public:
typedef void(CallerType::*CallerTypeFunc)();
function<void()>* constructFunction(CallerTypeFunc func)
{
CallerType* newMe = dynamic_cast<CallerType*> (this);
return new function<void()>(std::bind(func,newMe));
}
};
class A : public function<void()>
{
protected:
public:
A();
A(function<void()>* func) : function<void()>(*func)
{}
};
// now create ressource classes
// they provide functions to be called via an object of class A
class B : public AFunctionConstructor<B>
{
void foo()
{
cout << "Foo";
}
public:
A a;
B() : a(constructFunction(&B::foo)) {}
};
class C : public AFunctionConstructor < C >
{
void bar()
{
cout << "Bar";
}
public:
A a;
C() : a(constructFunction(&C::bar)) {}
};
int main()
{
B b;
C c;
b.a();
c.a();
cout << endl;
A* array[5];
array[0] = &b.a; //different functions with their ressources
array[1] = &c.a;
array[2] = &b.a;
array[3] = &c.a;
array[4] = &c.a;
for (int i = 0; i < 5; i++) //this usability i wanted to provide
{
(*(array[i]))();
}
getchar();
return 0;
}
Output :
FooBar
FooBarFooBarBar
This is as far as i can press it down concerning examples. But i guess this is unsafe code. I stumbled across possible other and simpler ways to achieve this (other uses of std::function and lambdas(which i might have tried to reinvent here partially it seems)).
At first I had tried to pass "this" to the bind function in function<void()>*AFunctionConstructor::constructFunction(CallerTypeFunc func)
,though, which i now get through the dynamic upcast.
Additionally the functionality of AFunctionConstructor was first supposed to be implemented in a Constructor of A.
I have 2 class, I would like to pass a method from one to other by callback!
See that I also wish to hold the address of this method using void (*callBack)();
I'm used to do this in C, but I dont know how to do this in c++;
#include <iostream>
using namespace std;
class A
{
private:
void (*callBack)(); //to hold the address of the method
public:
A();
void setCallBack(void(*cB)());
void useCallBack();
};
A::A()
{
}
void A::setCallBack(void(*cB)())
{
callBack = cB;
}
void A::useCallBack()
{
callBack();
}
class B
{
private:
A * Aguy;
public:
B();
void someMethod();
void otherMethod();
};
B::B()
{
Aguy = new A();
}
void B::otherMethod()
{
Aguy->setCallBack(someMethod);
Aguy->useCallBack()
}
void B::someMethod()
{
cout << "Hello. I'm from class b" << endl;
}
int main()
{
B Bguy;
Bguy.otherMethod();
return 0;
}
The problem is that:
void (*callBack)();
This is not a pointer to a method. This is a pointer to a function.
To have a pointer to a method you need to specify the class the method is in.
void (B::*callBack)();
Then when you call it you need to call it via an object.
void A::useCallBack(B* b)
{
(b->*callBack)();
}
But this is probably not what you want.
What you really want is a wrapper that encapsulates all this.
I would take a look at std::function. This will allow you to wrap a method call and an object into a single object that you can then call.
std::function<void()> callback;
Just replace all your occurrences of void(*cB)() with std::function<void()> then you can bind an instance of the object to the method at the call point.
Aguy->setCallBack(std::bind(&B::someMethod, this));
This also allows you to seemly pass any normal function or functor as a callback.
void print()
{ std:cout << "It worked\n";
}
...
Aguy->setCallBack(&print);
struct Printer
{
void operator()() const
{
std::cout << "It worked with obejct\n";
}
}
...
Aguy->setCallBack(Printer());
If you need to pass member function pointers see the modified code. it uses modern c++ constructs.
#include <iostream>
#include <functional>
using namespace std;
class A
{
private:
typedef std::function<void()> some_void_function_type;
some_void_function_type f_;
public:
A();
void setCallBack(some_void_function_type f);
void useCallBack();
};
A::A()
{
}
void A::setCallBack(some_void_function_type f)
{
f_ = f;
}
void A::useCallBack()
{
f_();
}
class B
{
private:
A * Aguy;
public:
B();
void someMethod();
void otherMethod();
};
B::B()
{
Aguy = new A();
}
void B::otherMethod()
{
Aguy->setCallBack(std::bind(&B::someMethod, this));
Aguy->useCallBack();
}
void B::someMethod()
{
cout << "Hello. I'm from class b" << endl;
}
int main()
{
B Bguy;
Bguy.otherMethod();
return 0;
}
See c++ - <unresolved overloaded function type> for details.
To quote the answer:
In C++, member functions have an implicit parameter which points to
the object (the this pointer inside the member function). Normal C
functions can be thought of as having a different calling convention
from member functions, so the types of their pointers
(pointer-to-member-function vs pointer-to-function) are different and
incompatible. C++ introduces a new type of pointer, called a
pointer-to-member, which can be invoked only by providing an object.
Put static on someMethod:
class B
{
private:
A * Aguy;
public:
B();
static void someMethod();
void otherMethod();
};
void B::otherMethod() {
Aguy->setCallBack(B::someMethod);
Aguy->useCallBack(); // adding missing semicolon
}
Is it possible in C++?
For example I have a pointer to a function that takes no parameters and its return type is void:
void (*f)();
and and a function object:
class A
{
public:
void operator()() { cout << "functor\n"; }
};
Is it possible to assign to f the address of an A object? And when I call f() to call the A functor?
I tried this but it doesn't work:
#include <iostream>
using namespace std;
class A
{
public:
void operator()() { cout << "functorA\n"; }
};
int main()
{
A ob;
ob();
void (*f)();
f = &ob;
f(); // Call ob();
return 0;
}
I get C:\Users\iuliuh\QtTests\functor_test\main.cpp:15: error: C2440: '=' : cannot convert from 'A *' to 'void (__cdecl *)(void)'
There is no context in which this conversion is possible
Is there any way to achieve this?
You can't do it in the way you've specified, because:
operator() must be a nonstatic function (standards requirement)
a pointer to a non-static function must have an implicit parameter - the pointer to the class instance
your call to f() does not give any indication on which instance of the object A your function is called
Using C++11 and std::function, as Stephane Rolland pointed out, may do the trick - you'll be specifying the pointer to the object in the binding:
std::function<void(void)> f = std::bind(&A::operator(), &ob);
(See question on using std::function on member functions)
If you use C++11, could use std::function
#include <functional>
std::function<void()> f;
int main()
{
A ob;
ob();
f = ob; // f refers to ob
f(); // Call ob();
return 0;
}
Yes it's kind of possible using a C++1/C++0x feature, but to achieve this you should use the std::function which can address to the two types, functions and object functions.
#include <functional>
class A
{
public:
void operator()() { }
};
int main()
{
std::function<void(void)> aFunction;
A ob;
aFunction = ob;
// or as another user said
// aFunction = std::bind(&A:operator(), &ob);
aFunction();
void (*f)();
aFunction = f;
aFunction();
return 0;
}
and if you're stuck with C++03, you could play with std::mem_fun and std::ptr_fun
How about some workaround like this:
Basically you want to have a common way of calling member functions and functions. Then maybe you could create a wrapper that would represent a generic pointer to either a function or member function. Let's say you have Base class and you want to be able to invoke operator() of all derived classes. Then you also have a function() that you want to invoke as well:
class Base
{
public:
virtual void operator()() = 0;
};
class A : public Base
{
public:
void operator()(){ std::cout << "A()" << std::endl; }
};
void function()
{
std::cout << "function" << std::endl;
}
If you create an wrapper that allows you to construct your custom pointer (MyFncPtr):
typedef void (Base::*BaseFncPtr)();
typedef void (*FncPtr)();
class MyFncPtr
{
public:
MyFncPtr(FncPtr f) : fnc(f), baseObj(NULL), baseFnc(NULL) { }
MyFncPtr(BaseFncPtr fPtr, Base* objPtr) : baseFnc(fPtr), baseObj(objPtr), fnc(NULL) { }
void invoke()
{
if (baseObj && baseFnc)
(baseObj->*baseFnc)();
else if (fnc)
fnc();
}
private:
BaseFncPtr baseFnc;
Base* baseObj;
FncPtr fnc;
};
you could achieve it like this:
A a;
MyFncPtr myPtr(&Base::operator(), &a);
myPtr.invoke();
MyFncPtr myPtr2(function);
myPtr2.invoke();
outputs:
A()
function
Hope this helps :)
There is a class
class A {
public:
A() {};
private:
void func1(int) {};
void func2(int) {};
};
I want to add a function pointer which will be set in constructor and points to func1 or func2.
So I can call this pointer (as class member) from every class procedure and set this pointer in constructor.
How can I do it?
class A {
public:
A(bool b) : func_ptr_(b ? &A::func1 : &A::func2) {};
void func(int i) {this->*func_ptr(i);}
private:
typedef void (A::*func_ptr_t_)();
func_ptr_t_ func_ptr_;
void func1(int) {};
void func2(int) {};
};
That said, polymorphism might be a better way to do whatever you want to do with this.
Add a member variable
void (A::*ptr)();
set it in the constructor
ptr=&A::func1;
(or use the initializer list) and call it in methods of A:
(this->*ptr)();
I compiled and ran this code. The various members need to be public so you can pass them into the constructor. Otherwise, here you go.
However, I agree with other posters that this is almost definitely a bad thing to do. ;) Just make invoke pure virtual, and then make two subclasses of A which each override invoke().
#include <iostream>
using namespace std;
class A;
typedef void(A::*MyFunc)(int) ;
class A {
public:
A() {}
A(MyFunc fp): fp(fp) {}
void invoke(int a)
{
(this->*fp)(a);
}
void func1(int a) { cout << "func1 " << a << endl; }
void func2(int a) { cout << "func2 " << a << endl; }
private:
MyFunc fp;
};
int main()
{
A* a = new A( & A::func1 );
a->invoke(5);
A* b = new A( & A::func2 );
b->invoke(6);
}
See boost::function for a way to handle function and class member pointers in a more OO/C++ manner.
For example (from the documentation) :
struct X
{
int foo(int);
};
boost::function<int (X*, int)> f;
f = &X::foo;
X x;
f(&x, 5);
I suggest you use functor(or function object), rather than function pointer, because the former is safer, and function pointer can be difficult or awkward to pass a state into or out of the callback function
A functor is basically a re-implementation of operator() of class A, for very detailed description please refer to Wikipedia: http://en.wikipedia.org/wiki/Function_object
The code should be something like this:
class A {
public:
A() {};
void operator()(int function_index, int parameter) {
if(function_index == 1)
func1(parameter);
else if(function_index == 2)
func2(parameter);
else
{ //do your other handling operation
}
}
private:
void func1( int ) {};
void func2( int) {};
};
By using that class:
A a;
a(1, 123); //calling func1
a(2, 321); //calling func2
Why do you think it's a bad thing to do. I just need one function pointer and I don't want to create two subclasses for this. So why is it so bad?
Some example...
class A; // forward declaration
typedef void (A::*func_type)(int);
class A {
public:
A() {
func_ptr = &A::func1;
}
void test_call(int a) {
(this->*func_ptr)(a);
}
private:
func_type func_ptr;
void func1(int) {}
void func2(int) {}
};