Call a pointer-to-function outside the structure - c++

I have a structure, inside it a pointer to function from the same structure. And now I need to call a pointer to function outside the structure. I give an example of the code below:
#include <iostream>
struct test {
void (test::*tp)(); // I need to call this pointer-to-function
void t() {
std::cout << "test\n";
}
void init() {
tp = &test::t;
}
void print() {
(this->*tp)();
}
};
void (test::*tp)();
int main() {
test t;
t.init();
t.print();
(t.*tp)(); // segfault, I need to call it
return 0;
}

(t.*tp)(); is trying to invoke the member function pointer tp which is defined at global namespace as void (test::*tp)();, note that it's initialized as null pointer in fact (via zero initialization1), invoking it leads to UB, anything is possible.
If you want to invoke the data member tp of t (i.e., t.tp) on the object t, you should change it to
(t.*(t.tp))();
^
|
---- object on which the member function pointed by tp is called
If you do want to invoke the global tp, you should initialize it appropriately, such as
void (test::*tp)() = &test::t;
then you can
(t.*tp)(); // invoke global tp on the object t
1 About zero initialization
Zero initialization is performed in the following situations:
1) For every named variable with static or thread-local storage duration that is not subject to constant initialization (since C++14), before any other initialization.

#songyuanyao's answer is valid. However, are you sure you want to use your structure that way? Why not just use inheritance and virtual methods? :
class base_test {
public:
virtual void t() { std::cout << "test\n"; }
void print() { t(); }
};
and then you can subclass it:
class my_test : base_test {
public:
virtual void t() { std::cout << "my test\n"; }
};
In your main() function (or wherever) you could have functions returning pointers or references to the base class, which are actually instances of subclasses. And this way, you don't have to worry about pointers.
The downside is that you have to know about your different tests at compile time (and then not even at the site of use, as I just explained). If you do, I'd go with the common idiom.

Related

Call methods of a std::is_empty type without an instance?

I am writing a template function, foo(), that takes a type T that is known to be empty (T has no members and std::empty<T>::value is true).
I want to call non-static member functions of this type without constructing any instances of it. Is this possible?
Here is some example code demonstrating what I am trying to do:
class Foo {
public:
int func() { return 4;}
};
template<typename T>
void example() {
// Call T::func() somehow without an instance of T?
}
int main() {
std::cout << example<Func>() << std::endl;
}
You've said
I don't control foo and can't change it to static.
Sounds like a bummer, since if you want to call an instance method, you need an instance, so either you go into UB, or you need to conjure the instance.
Since that T you say is 'empty', can you afford a single global shared instance, just for the purpose of having an instance? Then you can create a shared instance as local static:
//+out of reach
class Foo {
public:
int func() { return 4;}
};
class Bar {
public:
int func() { return 456;}
};
//-out of reach
template<typename T>
int example() {
static T shared_instance;
return shared_instance.func();
}
int main() {
std::cout << example<Foo>() << std::endl;
std::cout << example<Bar>() << std::endl;
}
However, if that class is really 'empty', I don't see why are you worried about something as simple as
template<typename T>
int example() {
T instance;
return instance.func();
}
Of course I cannot 100% guarantee, but I bet that any decent compiler will completely optimize out that instance under your specified conditions. Of course you should check it after compiling if you really need to rely on that function to have as small stack footprint as possible.

Pass member function pointer to parent class yields compiler error

I'd like to have child classes register callbacks to their parent class so that users of the parent class can call methods of the child with a known function signature.
typedef int(*Func)(int);
class A
{
public:
void registerFunc(Func f)
{}
};
class B : public A
{
public:
B()
{
A::registerFunc(&B::myF);
}
int myF(int x) {
// do stuff with member variables
return 3;
}
};
But I get this compiler error
main.cpp:18:23: error: cannot initialize a parameter of type 'Func' (aka 'int (*)(int)') with an rvalue of type 'int (B::*)(int)'
A::registerFunc(&B::myF);
^~~~~~~
main.cpp:8:28: note: passing argument to parameter 'f' here
void registerFunc(Func f)
Here's a Repl illustrating the error in a concise example.
https://replit.com/#Carpetfizz/RudeSmoothComments#main.cpp
The accepted answer in a related thread suggested to override a virtual function declared in A but my use case actually requires dynamic callback registrations.
You can try this.
typedef std::function<int (int)> Func;
class A
{
public:
void registerFunc(Func f)
{}
};
class B : public A
{
public:
B()
{
A::registerFunc(std::bind(&B::myF, *this, std::placeholders::_1));
}
int myF(int x) {
// do stuff with member variables
return 3;
}
};
If I understand the goal (and believe me, that's a sketchy 'if'), you want to specify some member of some A derivation to invoke from some A member as a dispatched 'callback' mechanic. If that is the case, then to answer your question in comment, yes, a function and bind can do this. It can even be semi-protected with a little help from sfinae:
Example
#include <iostream>
#include <type_traits>
#include <functional>
#include <memory>
struct A
{
virtual ~A() = default;
std::function<void(int)> callback = [](int){};
template<class Derived>
std::enable_if_t<std::is_base_of<A, Derived>::value>
registerCallback(void (Derived::*pfn)(int))
{
using namespace std::placeholders;
callback = std::bind(pfn, dynamic_cast<Derived*>(this), _1);
}
void fire(int arg)
{
callback(arg);
}
};
struct B : public A
{
void memberfn(int arg)
{
std::cout << __PRETTY_FUNCTION__ << ':' << arg << '\n';
}
};
struct Foo
{
void memberfn(int arg)
{
std::cout << __PRETTY_FUNCTION__ << ':' << arg << '\n';
}
};
int main()
{
std::unique_ptr<A> ptr = std::make_unique<B>();
ptr->registerCallback(&B::memberfn);
// ptr->registerCallback(&Foo::memberfn); // WILL NOT WORK
ptr->fire(42);
}
Output
void B::memberfn(int):42
The Parts
The first part is straight forward. We declare a member variable callback to be a std::function<void(int)> instance. This is where we'll eventually bind our callable object point. The default value is a lambda that does nothing.
The second part is... a little more complicated:
template<class Derived>
std::enable_if_t<std::is_base_of<A, Derived>::value>
registerCallback(void (Derived::*pfn)(int))
This declares registerCallback as an available member function that accepts a non-static member function pointer taking one int as an argument, but only if the class hosting that member function, or a derivative therein, is a derivation of A (or A itself). Some non-A derivative Foo with a member void foo(int) will not compile.
Next, the setup to the callback itself.
using namespace std::placeholders;
callback = std::bind(pfn, dymamic_cast<Derived*>(this), _1);
This just binds the pointer-to-member to this dynamic-cast to the derivation type (which had better work or we're in trouble, see final warning at the end of this diatribe), and sets the call-time placeholder. The _1 you see comes from the std::placeholders namespace, and is used to delay providing an argument to the callback until such time as we actually invoke it (where it will be required,and you'll see that later). See std::placehholders for more information.
Finally, the fire member, which does this:
void fire(int arg)
{
callback(arg);
}
This invokes the registered function object with the provided argument. Both the member function and this are already wired into the object. The argument arg is used to fill in the placeholder we mentioned earlier.
The test driver for this is straightforward:
int main()
{
std::unique_ptr<A> ptr = std::make_unique<B>();
ptr->registerCallback(&B::memberfn);
// ptr->registerCallback(&Foo::memberfn); // WILL NOT WORK
ptr->fire(42);
}
This creates a new B, hosting it in a dynamic A pointer (so you know there is no funny business going on). Even with that, because B derived from A the registerCallback sfinae filtering passes inspection and the callback is registered successfully. We then invoke the fire method, passing our int argument 42, which will be sent to the callback, etc.
Warning: With great power comes great responsibility
Even those there is protection from passing non-A derived member functions, there is absolutely none from the casting itself. It would be trivial to craft a basic A, pass a B member (which will work since A is its base), but there is no B actually present.
You can catch this at runtime via that dynamic_cast, which we're currently not error checking. For example:
registerCallback(void (Derived::*pfn)(int))
{
using namespace std::placeholders;
Derived *p = dynamic_cast<Derived*>(this);
if (p)
callback = std::bind(pfn, p, _1);
}
You can choose the road more risky. Personally, i'd detect the null case and throw an exception just to be safe(er)

C++ invalid function type casting

I've read several topics about that kind of problem - but can't find a simple and good solution. Here is the code:
void SomeFunction() { }
class A {
public:
typedef std::function<void(void)> AFunction;
static void AMethod(AFunction f) { f(); }
};
class B {
public:
void B1Method() { }
void BCorrectCall() { A::AMethod(SomeFunction); }
void BIncorrectCall() { A::AMethod(B1Method); }
};
Problem is here void BIncorrectCall() { A::AMethod(B1Method); }, where I receive error about invalid casting. What is the simplest way to achieve that kind of behaviour? Thanks a lot for any advice!
Use a lambda:
A::AMethod([this]{B1Method();});
It doesn't matter in this case, but if you wanted to store AFunction f and use it after the call to AMethod, you'd have to ensure that the B instance (the address of which is saved in the lambda) says alive as long as you use the function.
C++17 allows you to capture *this instead, which will copy the entire B instance into lambda, but normally it's not what you want.
You could do something similar with std::bind (see the other answer), but lambdas are more flexible.
B1Method is not void(*)(void), it's void(B1::*)(void).
You may do
void BIncorrectCall() { A::AMethod(std::bind(&B1::B1Method, this)); }
};
The issue is that B::B1Method() is a non-static member function in B and, therefore, it needs to be called on an instance of B.
If the implementation of B1Method() doesn't use any non-static data member of B and it doesn't call any other non-static member function of B, then simply declaring it as static will work with your current implementation of BIncorrectCall() as you will no longer need to call B1Method() on an instance of B:
class B {
public:
static void B1Method() { } // static now
void BCorrectCall() { A::AMethod(SomeFunction); }
void BIncorrectCall() { A::AMethod(B1Method); } // no change
};
Otherwise, you have to keep an object of type B whenever you want to call B1::B1Method().
The easiest way is to make it static and so there is no this object, but if you need it (the this object), you can use lambdas:
class B {
public:
void B1Method() { }
void BCorrectCall() { A::AMethod(SomeFunction); }
void BIncorrectCall() {
std::function<void(void)> el = [&](){this->B1Method();};
A::AMethod(el);
}
};
The problem is that 'B1Method' is not a simple function - it's a class method. That means that when you call myB.B1Method(), you're actually calling 'B1Method(&myB)', effectively passing the this pointer as a hidden argument - so you can't convert M1Method to a std::function without specifying which object it should act on.
One approach that should work is using std::bind to construct a callable object from a combination of an object (class instance) and the method. Something like:
void BNowCorrectCall() { A::AMethod(std::bind(&B::B1Method, this)); }

C++ How to make function pointer to class method [duplicate]

This question already has answers here:
Calling C++ member functions via a function pointer
(10 answers)
Closed 9 months ago.
I'm having trouble making a function pointer to a class method. I made a function pointer to a non-class method and it works fine.
int foo(){
return 5;
}
int main()
{
int (*pointer)() = foo;
std::cout << pointer();
return 0;
}
I tried to apply this to have an instance variable in a class be a function pointer.
This is the header file. It declares the private method Print which the variable method will point to.
class Game
{
public:
Game();
private:
void Print();
void (method)( void );
};
The Game constructor attempts to assign the pointer method to the address of the Print method. Upon compile, an error comes up at that line saying "error: reference to non-static member function must be called;". I don't know what that means. Whats the correct way of implementing this?
Game::Game( void )
{
method = &Game::Print;
}
void Game::Print(){
std::cout << "PRINT";
}
A member function is quite a bit different from an ordinary function, so when you want to point to a member function you need a pointer-to-member-function, not a mere pointer-to-function. The syntax for a pointer-to-member-function includes the class that the member function is a member of:
void (Game::*mptr)();
This defines a pointer-to-member-function named mptr that holds a pointer to a member function of the class Games that takes no arguments and returns nothing. Contrast that with an ordinary function pointer:
void (*ptr)();
This defined a pointer-to-function named ptr that holds a pointer to a function that takes no arguments and returns nothing.
Just found out you can do
#include <functional>
#include <cassert>
using namespace std;
struct Foo {
int x;
int foo() { return x; }
};
int main() {
function<int(Foo&)> f = &Foo::foo;
Foo foo = { 3 };
assert(f(foo) == 3);
foo.x = 5;
assert(f(foo) == 5);
}
std::mem_fn() might work too: https://en.cppreference.com/w/cpp/utility/functional/mem_fn
1- Use the following syntax to point to a member function:
return_type (class_name::*ptr_name) (argument_type) = &class_name::function_name;
2- keep in mind to have a pointer to member function you need to make it public.
in your case:
class Game
{
public:
Game(){}
void print() {
//todo ...
}
};
// Declaration and assignment
void (Game::*method_ptr) () = &Game::print;

The use case of 'this' pointer in C++

I understand the meaning of 'this', but I can't see the use case of it.
For the following example, I should teach the compiler if the parameter is the same as member variable, and I need this pointer.
#include <iostream>
using namespace std;
class AAA {
int x;
public:
int hello(int x) { this->x = x;}
int hello2(int y) {x = y;} // same as this->x = y
int getx() {return x;}
};
int main()
{
AAA a;
a.hello(10); // x <- 10
cout << a.getx();
a.hello2(20); // x <- 20
cout << a.getx();
}
What would be the use case for 'this' pointer other than this (contrived) example?
Added
Thanks for all the answers. Even though I make orangeoctopus' answer as accepted one, it's just because he got the most vote. I must say that all the answers are pretty useful, and give me better understanding.
Sometimes you want to return yourself from an operator, such as operator=
MyClass& operator=(const MyClass &rhs) {
// assign rhs into myself
return *this;
}
The 'this' pointer is useful if a method of the class needs to pass the instance (this) to another function.
It's useful if you need to pass a pointer to the current object to another function, or return it. The latter is used to allow stringing functions together:
Obj* Obj::addProperty(std::string str) {
// do stuff
return this;
}
obj->addProperty("foo")->addProperty("bar")->addProperty("baz");
In C++ it is not used very often. However, a very common use is for example in Qt, where you create a widget which has the current object as parent. For example, a window creates a button as its child:
QButton *button = new QButton(this);
When passing a reference to an object within one of its methods. For instance:
struct Event
{
EventProducer* source;
};
class SomeContrivedClass : public EventProducer
{
public:
void CreateEvent()
{
Event event;
event.source = this;
EventManager.ProcessEvent(event);
}
};
Besides obtaining a pointer to your own object to pass (or return) to other functions, and resolving that an identifier is a member even if it is hidden by a local variable, there is an really contrived usage to this in template programming. That use is converting a non-dependent name into a dependent name. Templates are verified in two passes, first before actual type substitution and then again after the type substitution.
If you declare a template class that derives from one of its type parameters you need to qualify access to the base class members so that the compiler bypasses the verification in the first pass and leaves the check for the second pass:
template <typename T>
struct test : T {
void f() {
// print(); // 1st pass Error, print is undefined
this->print(); // 1st pass Ok, print is dependent on T
}
};
struct printer {
void print() { std::cout << "print"; }
};
struct painter {
void paint() { std::cout << "paint"; }
};
int main() {
test<printer> t; // Instantiation, 2nd pass verifies that test<printer>::print is callable
t.f();
//test<painter> ouch; // 2nd pass error, test<painter>::print does not exist
}
The important bit is that since test inherits from T all references to this are dependent on the template argument T and as such the compiler assumes that it is correct and leaves the actual verification to the second stage. There are other solutions, like actually qualifying with the type that implements the method, as in:
template <typename T>
struct test2 : T {
void f() {
T::print(); // 1st pass Ok, print is dependent on T
}
};
But this can have the unwanted side effect that the compiler will statically dispatch the call to printer::print regardless of whether printer is a virtual method or not. So with printer::print being declared virtual, if a class derives from test<print> and implements print then that final overrider will be called, while if the same class derived from test2<print> the code would call printer::print.
// assumes printer::print is virtual
struct most_derived1 : test<printer> {
void print() { std::cout << "most derived"; }
};
struct most_derived2 : test2<printer> {
void print() { std::cout << "most derived"; }
};
int main() {
most_derived1 d1;
d1.f(); // "most derived"
most_derived2 d2;
d2.f(); // "print"
}
You can delete a dynamically created object by calling delete this from one of its member functions.
The this pointer is the pointer to the object itself. Consider for example the following method:
class AAA {
int x;
public:
int hello(int x) { some_method(this, x);}
};
void somefunc(AAA* a_p)
{
......
}
class AAA {
int x;
public:
int hello(int x) { this->x = x;}
int hello2(int y) {x = y;} // same as this.x = y
int getx() {return x;}
void DoSomething() { somefunc(this); }
};
this is implicit whenever you use a member function or variable without specifying it. Other than that, there are many, many situations in which you'll want to pass the current object to another function, or as a return value.
So, yeah, it's quite useful.
Sometimes you need to refer to "this" object itself, and sometimes you may need to disambiguate in cases where a local variable or a function parameter shadows a class member:
class Foo {
int i;
Foo* f() {
return this; // return the 'this' pointer
}
void g(){
j(this); // pass the 'this' pointer to some function j
}
void h(int i) {
this->i = i; // need to distinguish between class member 'i' and function parameter 'i'
}
};
The two first cases (f() and g() are the most meaningful cases. The third one could be avoided just by renaming the class member variable, but there's no way around using this in the first two cases.
Another possible use case of this:
#include <iostream>
using namespace std;
class A
{
public:
void foo()
{
cout << "foo() of A\n";
}
};
class B : A
{
public:
void foo()
{
((A *)this)->foo(); // Same as A::foo();
cout << "foo() of B\n";
}
};
int main()
{
B b;
b.foo();
return 0;
}
g++ this.cpp -o this
./this
foo() of A
foo() of B
One more use of this is to prevent crashes if a method is called on a method is called on a NULL pointer (similar to the NULL object pattern):
class Foo
{
public:
void Fn()
{
if (!this)
return;
...
}
};
...
void UseFoo(Foo* something)
{
something->Fn(); // will not crash if Foo == NULL
}
If this is useful or not depends on the context, but I've seen it occasionally and used it myself, too.
self-assignment protection