In C ++, I first encountered function pointers.
I tried to use this to make it similar to Action and Delegate in C #.
However, when declaring a function pointer, it is necessary to specify the type of the class in which the function exists.
ex) void (A :: * F) ();
Can I use a function pointer that can store a member function of any class?
In general, function pointers are used as shown in the code below.
class A {
public:
void AF() { cout << "A::F" << endl; }
};
class B {
public:
void(A::*BF)();
};
int main()
{
A a;
B b;
b.BF = &A::AF;
(a.*b.BF)();
return 0;
}
I want to use it like the code below.
is this possible?
Or is there something else to replace the function pointer?
class A {
public:
void AF() { cout << "A::F" << endl; }
};
class B {
public:
void(* BF)();
};
int main()
{
A a;
B b;
b.BF = a.AF;
return 0;
}
I solved the question through the answer.
Thanks!
#include <functional>
#include <iostream>
class A {
public:
void AF() { std::cout << "A::F" << std::endl; }
};
class C {
public:
void CF() { std::cout << "C::F" << std::endl; }
};
class B {
public:
B(){}
std::function<void()> BF;
};
int main() {
A a;
C c;
B b;
b.BF = std::bind(&A::AF, &a);
b.BF();
b.BF = std::bind(&C::CF, &c);
b.BF();
int i;
std::cin >> i;
return 0;
}
What you want to do is probably something like this. You can use std::function to hold a pointer to a member function bound to a specific instance.
#include <functional>
#include <iostream>
class A {
public:
void AF() { std::cout << "A::F" << std::endl; }
};
class B {
public:
B(const std::function<void()>& bf) : BF(bf) {}
std::function<void()> BF;
};
int main() {
A a;
B b1(std::bind(&A::AF, &a)); // using std::bind
B b2([&a] { a.AF(); }); // using a lambda
b1.BF();
b2.BF();
return 0;
}
Here's a C# style implementation of the accepted answer, It is memory efficient and flexible as you can construct and delegate at different points of execution which a C# developer might expect to do:
#include <iostream>
#include <functional>
using namespace std;
class A {
public:
void AF() { cout << "A::F" << endl; }
void BF() { cout << "B::F" << endl; }
};
class B {
public:
std::function<void()> Delegate;
};
int main() {
A a;
B b;
b.Delegate = std::bind(&A::AF, &a);
b.Delegate();
b.Delegate = [&a] { a.BF(); };
b.Delegate();
return 0;
}
Related
as the below code, I don't want so many "if else"
class A
{
public:
void f0()
{
cout << "f0" << endl;
}
void f1()
{
cout << "f1" << endl;
}
void f2()
{
cout << "f2" << endl;
}
//..... more functions fn()...
};
class B
{
public:
void f(int n)
{
//vector< function<void()> > f_v {obj_a.f0, obj_a.f1, obj_a.f2}; //this usage is not correct
if (n == 0)
obj_a.f0();
else if (n == 1)
obj_a.f1();
else if (n == 2)
obj_a.f2();
//.....more else if here
}
private:
A obj_a;
};
I want to create a vector and use std::function to avoid using so many if-else , like vector< function<void()> > f_v {obj_a.f0, obj_a.f1, obj_a.f2}; but it doesn't work, maybe the usage of std::function is not very correct. How should I do? Or is there any other good way to solve the problem of if else, I think use so many switch-case is also not very elegant :)
update:
Some answers have already solve my problem about the usage of std::function in my earlier code;
More generally, considering the below code, if the member functons A::f1(), A::f2().... have different return types, but still have some connection that they derived from a same base class , what's the good way to implement the logic of if else in B::f()?
class Base
{
public:
virtual ~Base()=default;
};
class D1 : public Base
{
public:
};
class D2 : public Base
{
public:
};
class D3 : public Base
{
public:
};
// ....maybe more class derived form Base
class A
{
public:
D1* f0()
{
cout << "f0" << endl;
return &d1;
}
D2* f1()
{
cout << "f1" << endl;
return &d2;
}
D3* f2()
{
cout << "f2" << endl;
return &d3;
}
//more fn()....
private:
D1 d1;
D2 d2;
D3 d3;
//.....
};
class B
{
public:
void f(int n)
{
if (n == 0)
obj_a.f0();
else if (n == 1)
obj_a.f1();
else if (n == 2)
obj_a.f2();
//.....more else if here
}
private:
A obj_a;
};
You can use std::function with a lambda wrapper,
vector<function<void()>> f_v {[this]() { obj_a.f0(); },
[this]() { obj_a.f1(); },
[this]() { obj_a.f2(); }};
f_v[n]();
or use pointer-to-members directly,
vector<void (A::*)()> f_v { &A::f0, &A::f1, &A::f2 };
(obj_a.*f_v[n])();
If you are aiming at speed and you know the number of methods, avoid using std::vector for the extra indirection. Use std::array as it will hit cache with the current object.
For this simple case, you don't necessarily need to use std::function, which is a very heavy object to call. You can use pointers to members like this:
#include <iostream>
#include <array>
using namespace std;
class A
{
public:
void f0()
{
cout << "f0" << endl;
}
void f1()
{
cout << "f1" << endl;
}
void f2()
{
cout << "f2" << endl;
}
//.....
};
class B
{
public:
B() {
fn[0] = &A::f0;
fn[1] = &A::f1;
fn[2] = &A::f2;
}
void f(int n)
{
((obj_a).*(fn[n]))();
}
private:
using Fn = void (A::*)();
std::array<Fn,3> fn;
A obj_a;
};
Code: https://godbolt.org/z/z4KqKvn99
Your approach seems correct. You just need to std::bind those member functions like
class B{
std::vector <std::function <void()>> m_vec_functs;
...
B(const A& a)
{
m_vec_functs.push_back(std::bind(&A::f0, a));
m_vec_functs.push_back(std::bind(&A::f1, a));
}
void f(unsigned int n)
{
m_vec_functs[n];
}
...
};
Then you can access each individual function by its index, knowing their order.
Looking for how to best access class B's queue through A but I am receiving a segmentation fault. Also I am looking for the best way to communicate between these two classes. Are accessor methods ok in this scenario? What design pattern could work? Thanks
class B {
public:
int get_int() { return qi.front(); }
void put_int(int i) { qi.push(i); }
private:
queue<int> qi;
};
class A
{
public:
void get_event() { cout << b->get_int() << endl; }
void put_event(int a) { b->put_int(a); }
private:
B *b;
};
int main() {
A a;
a.put_event(1);
return 0;
}
As mentioned in comment problem is undefined initialization
you can fix that by using constructor for initialization
#include<iostream>
#include<queue>
using namespace std;
class B {
public:
int get_int() { return qi.front(); }
void put_int(int i)
{
qi.push(i);
}
private:
queue<int> qi;
};
class A
{
public:
void get_event() { cout << b->get_int() << endl; }
void put_event(int a) { b->put_int(a); }
A()
{
b = new B();
}
~A() { delete b; }
private:
B *b;
};
int main() {
A a;
a.put_event(1);
a.get_event();
return 0;
}
Output
1
Program ended with exit code: 0
A a;
is an undefined reference, you have to initialize it with a costructor and since you didn't defined any, you must use the default one
A a=new A();
or better, write the costructors of the two classes as you prefer and use them.
I got quite confused about the output of the following code. The output is
A::Fun
C::Do
Could anyone explain why this happened? Any help would be appreciated!
#include <iostream>
using namespace std;
class A {
private:
int nVal;
public:
void Fun()
{ cout << "A::Fun" << endl; }
void Do()
{ cout << "A::Do" << endl; }
};
class B:public A {
public:
virtual void Do()
{ cout << "B::Do" << endl; }
};
class C:public B {
public:
void Do( )
{ cout << "C::Do" <<endl; }
void Fun()
{ cout << "C::Fun" << endl; }
};
void Call(B & p) {
p.Fun(); p.Do();
}
int main() {
C c; Call(c);
return 0;
}
The Fun function is not virtual in any base-class. In the function Call all the compiler knows about is the A::Fun function, it doesn't know about the C class. All it knows is that you have a reference to a B object, and there is no B::Fun so it looks in the parent class and find the A::Fun function.
Is there any way in C++ to collect different types of classes and call their methods?
What I want to do is as below,
template<namespace T>
class A
{
A method_A1(T a)
{
...
}
void method_A2(int aa)
{
...
}
...
};
class B
{
...
};
class C
{
...
};
class D
{
...
};
A<B> *b;
A<C> *c;
A<D> *d;
b -> method_A2(3);
c -> method_A2(5);
In this code object b,c,d they are totally different object, right? Not related.
But I want to bind them with a array, so...
z[0] = b;
z[1] = c;
z[2] = d;
like this.
I found some solutions, however the solutions are only for collecting different types. (using void* arrays or vectors for inherited objects) I also wanna access to their methods.
z[0] -> method_A2(3);
z[1] -> method_A3(5);
like this.
In this case how should I do?
Thanks in advance.
typedef boost::variant<A<B>, A<C>, A<D>> AVariant;
std::array<AVariant, 3> z;
z[0] = *b;
z[1] = *c;
z[2] = *d;
Then you can inspect each element's type if needed, or "visit" them using boost::static_visitor as shown here: http://www.boost.org/doc/libs/release/doc/html/variant.html
Why you don't use inheritance and polymorphisms. I have posted an example of what exactly can be a solution of your problem. See the main function:
#include <iostream>
class weapon {
public:
int fireRate;
int bulletDamage;
int range;
int activeBullet;
public:
virtual void fire(void) {std::cout << "machine " << '\n';}
virtual ~weapon() {std::cout << "destructor is virtual" << '\n';}
};
class machineGun: public weapon {
public:
void fire(void) {std::cout << "machine gun firing" << '\n';}
~machineGun(void) { std::cout << "machine gun destroyed" << '\n';}
};
class flamer: public weapon {
public:
void fire(void) {std::cout << "flamer firing" << '\n';}
~flamer(void) {std::cout << "flamer destroyed" << '\n';}
};
int main(void)
{
const int count = 2;
weapon *weapons[count];
machineGun *a = new machineGun();
flamer *b = new flamer();
weapons[0] = a;
weapons[1] = b;
weapons[0]->fire();
weapons[1]->fire();
delete a;
delete b;
}
If you don't want to change your classes' hierarchy, you can try having an array of callable objects. Something like:
#include <iostream>
#include <functional>
#include <array>
class A
{
public:
void Foo(int a)
{
std::cout << "Foo " << a << std::endl;
}
};
class B
{
public:
void Bar(int a)
{
std::cout << "Bar " << a << std::endl;
}
};
int main()
{
using namespace std::placeholders;
A a;
B b;
auto a_func = std::bind(&A::Foo, a, _1);
auto b_func = std::bind(&B::Bar, b, _1);
std::array<std::function<void(int)>, 2> arr = {
std::bind(&A::Foo, a, _1),
std::bind(&B::Bar, b, _1)
};
arr[0](1);
arr[1](2);
return 0;
}
BTW, this will only work if you use compiler with full C++11 support.
i'd like to invoke runtime-bound functions of classes, that inherit a binding ability from a common class "Bindable". Is that actually possible?
Here's a stub which surely lacks a lot of template-arguments and namespaces:
#include <iostream> // std::cout
#include <functional> // std::bind
#include <map> // std::map
class Bindable {
public:
void bindFunction (int x, auto newFn) {
mFns.insert(std::pair<int, auto>(x,newFn));
}
void invokeFunction (int key) {
mFns.at(key)();
}
protected:
std::map<int, function> mFns;
};
class A : Bindable {
void funAone (void) {
cout << "called funAone" <<std::endl;
}
void funAtwo (void) {
cout << "called funAtwo" <<std::endl;
}
};
class B : Bindable {
void funBone (void) {
cout << "called funBone" <<std::endl;
}
void funBtwo (void) {
cout << "called funBtwo" <<std::endl;
}
};
int main() {
A a;
B b;
a.bindFunction(1, &A::funAone);
a.bindFunction(2, &A::funAtwo);
b.bindFunction(1, &B::funBone);
b.bindFunction(2, &B::funBtwo);
a.invokeFunction(1);
a.invokeFunction(2);
b.invokeFunction(1);
b.invokeFunction(2);
}
Option #1
Use a CRTP idiom to know what type of pointers to member functions can be stored:
template <typename T>
struct Bindable {
void bindFunction (int x, void(T::*newFn)()) {
mFns.insert(std::make_pair(x,newFn));
}
void invokeFunction (int key) {
(static_cast<T*>(this)->*mFns.at(key))();
}
protected:
std::map<int, void(T::*)()> mFns;
};
struct A : Bindable<A> {
void funAone (void) {
std::cout << "called funAone" <<std::endl;
}
void funAtwo (void) {
std::cout << "called funAtwo" <<std::endl;
}
};
DEMO 1
Option #2
Use a type-erasure and make bindFunction a function template:
struct Bindable {
template <typename T, typename std::enable_if<std::is_base_of<Bindable, T>{}, int>::type = 0>
void bindFunction (int x, void(T::*newFn)()) {
mFns.insert(std::make_pair(x, std::bind(newFn, static_cast<T*>(this))));
}
void invokeFunction (int key) {
mFns.at(key)();
}
protected:
std::map<int, std::function<void()>> mFns;
};
struct A : Bindable {
void funAone (void) {
std::cout << "called funAone" <<std::endl;
}
void funAtwo (void) {
std::cout << "called funAtwo" <<std::endl;
}
};
DEMO 2
In both cases you can use the code as follows:
int main() {
A a;
B b;
a.bindFunction(1, &A::funAone);
a.bindFunction(2, &A::funAtwo);
b.bindFunction(1, &B::funBone);
b.bindFunction(2, &B::funBtwo);
a.invokeFunction(1);
a.invokeFunction(2);
b.invokeFunction(1);
b.invokeFunction(2);
}
Output:
called funAone
called funAtwo
called funBone
called funBtwo
Yes, it's possible, using std::bind. Note that auto can't be used as a function or template argument.
#include <iostream> // std::cout
#include <functional> // std::bind
#include <map> // std::map
class Bindable {
public:
typedef std::function<void()> Function;
void bindFunction (int x, Function newFn) {
mFns.insert(std::pair<int, Function>(x,newFn));
}
void invokeFunction (int key) {
mFns.at(key)();
}
protected:
std::map<int, Function > mFns;
};
class A : public Bindable {
public:
void funAone (void) {
std::cout << "called funAone" <<std::endl;
}
void funAtwo (void) {
std::cout << "called funAtwo" <<std::endl;
}
};
class B : public Bindable {
public:
void funBone (void) {
std::cout << "called funBone" <<std::endl;
}
void funBtwo (void) {
std::cout << "called funBtwo" <<std::endl;
}
};
int main() {
A a;
B b;
a.bindFunction(1, std::bind(&A::funAone, a)); // more than one way to bind
a.bindFunction(2, std::bind(&A::funAtwo, &a)); // the object parameter
b.bindFunction(1, std::bind(&B::funBone, b));
b.bindFunction(2, std::bind(&B::funBtwo, &b));
a.invokeFunction(1);
a.invokeFunction(2);
b.invokeFunction(1);
b.invokeFunction(2);
}