Create functions in preproccesor in C++ - c++

I'm using a timer library in Arduino IDE that has a function which takes "void(*func)(void)" as a parameter and runs it after certain time.I need to pass a function in my code with that library, the problem is I need "n" functions that do the same thing (just the variables involved inside the function take different values, which I take from an array of struct), that's why I can't use only one function and I need to be able to compile different functions with identical syntax but different variables involved.
I don't know how I can resolve this, what I want to do is to be able to create functions like in a for loop,
for(int i=0;i<MAX_FUNC;i++)
{
void func[i](){
x= structuredata.x[i];
}
}
I know this syntax is completely wrong but I'd like to know how can I achieve something like this . I tried using classes instead, but the library didn't let me use void functions created in a class as parameters, it indicated "unresolved overloaded function type"

If you're into template overrides and can restructure your functions as such, a solution to packing up a stack of callbacks to be fired consecutively is closer than you may think. You can make this more generic if the consistency of the function is literally identical. I.e. you can do this instead:
#include <iostream>
int ar[5] = { 1,2,3,4,5 };
// your function here
template <int i>
void callback()
{
// your code here. i is provided
int x = ar[i];
std::cout << "Got Value: " << x << '\n';
};
////////////////////////////////////////////////////////////
// unraveller
template<int N>
void callback_set()
{
callback_set<N-1>();
callback<N-1>();
}
// base callback set
template<>
void callback_set<0>() {}
////////////////////////////////////////////////////////////
int main()
{
callback_set<3>();
return 0;
}
Output
Got Value: 1
Got Value: 2
Got Value: 3
You "register" callback_set<N> as your callback, and it will chain up the calls to your function(s) for you. If individualized access to different implementations is needed, you can still to that, using full specializations:
#include <iostream>
// prototypes
template <int i>
void callback() {};
// your functions here
template<>
void callback<0>()
{
std::cout << "Callback #0" << '\n';
}
template<>
void callback<1>()
{
std::cout << "Callback #1" << '\n';
}
template<>
void callback<2>()
{
std::cout << "Callback #2" << '\n';
}
////////////////////////////////////////////////////////////
// unraveller
template<int N>
void callback_set()
{
callback_set<N-1>();
callback<N-1>();
}
// base callback set
template<>
void callback_set<0>() {}
////////////////////////////////////////////////////////////
int main()
{
callback_set<3>();
return 0;
}
Output
Callback #0
Callback #1
Callback #2
In essence you style-up your registered callback using callback_set<N> where N is the number of callbacks you've provided overrides for. Just a thought, but seems like it would be a fairly elegant solution to your problem if I understand it correctly.

You need some metaprogramming for that...
const int MAX_FUNC = 1000;
typedef void (*Tcallback)();
struct Tdata
{
int whatever;
};
Tdata callbackData[MAX_FUNC];
Tcallback callbacks[MAX_FUNC];
template <int index>
void callbackToIndex()
{
// callback knows it's index and thus can access some global structure for it's data
printf("%u", index);
callbackData[index];
}
template <int index>
void callbackCreate()
{
callbacks[index] = &callbackToIndex<index>;
callbackCreate<index - 1>();
}
template <>
void callbackCreate<-1>()
{
}
void create()
{
callbackCreate<MAX_FUNC - 1>();
}

Related

C++: use void as template argument

I have this minimal class to represent an event which client can subscribe to.
The event can have an data type associated to it, so when it is triggered by a publisher, an argument of that type would be passed to the client's callback:
template<typename Arg, typename Callback = function<void(const Arg&)>>
class Event
{
public:
Event(Callback c) : mCallback(c){}
void Trigger(const Arg& arg) {
mCallback(arg);
}
private:
Callback mCallback;
};
Now I can create an Event<int> or any other concrete type, but it is really important to me to also allow "empty" event, which has no data associated with it: Event<void>
But sadly that doesn't work:
static void FooVoid() {
cout << "Look ma, no args!" << endl;
}
static void FooInt(int a) {
cout << "int arg " << a << endl;
}
int main()
{
/* Compiles */
Event<int> eInt(&FooInt);
eInt.Trigger(42);
/* Does not compile :(
Event<void> eVoid(&FooVoid);
eVoid.Trigger();
*/
return 0;
}
Is there any way to achieve this desired API? How?
(P.S the solution should work on C++11)
The quickest way of solving this without explicitly specializing for void is to use a parameter pack (added in C++11) for your template argument instead of a single type and using an empty parameter pack instead of void. A parameter pack can homogeneously hold any number of type, including 0 and 1. Then it can be used to generate the right types and member functions. You basically just have to add ... correctly near every use of Arg (link) :
#include <functional>
#include <iostream>
template<typename ... Arg>
class Event
{
public:
using Callback = std::function<void(const Arg&...)>;
Event(Callback c) : mCallback(c){}
void Trigger(const Arg& ... arg) {
mCallback(arg...);
}
private:
Callback mCallback;
};
static void FooVoid() {
std::cout << "Look ma, no args!" << std::endl;
}
static void FooInt(int a) {
std::cout << "int arg " << a << std::endl;
}
int main()
{
/* Compiles */
Event<int> eInt(&FooInt);
eInt.Trigger(42);
Event<> eVoid(&FooVoid);
eVoid.Trigger();
return 0;
}
This has the added benefit that you can use callbacks with more than one argument. If this isn't desirable you can add a static_assert to prevent it :
template<typename ... Arg>
class Event
{
public:
using Callback = std::function<void(const Arg&...)>;
static_assert(sizeof...(Arg) <= 1, "Too many arguments");
Event(Callback c) : mCallback(c){}
void Trigger(const Arg& ... arg) {
mCallback(arg...);
}
private:
Callback mCallback;
};
Notice that this solution requires Event<> instead of Event<void>. You can solve that by adding a short specialization for Event<void> that uses Event<> (link) :
template<>
class Event<void> : public Event<>
{
// Inherit constructors
using Event<>::Event;
};

using pointer to inline function vs using pointer to function

I want to pass pointers of some functions to a template class to use them later. I was wondering if:
Does it make a possibly (speed-wise) beneficial difference if I make these functions inline?
Functions themselves are possibly one line wrapper for another functions like the example below:
//inline ?
void func_wrapper_1(){
func1(arg1);
}
//inline ?
void func_wrapper_2(){
func2(arg2);
}
and the class template is like the example below:
template<void(*f1)(), void(*f2)()>
class caller{
public:
static void func(int v){
if(v) {
(*f1)();
}else{
(*f2)();
}
}
};
And later on in the main function it will be used like the example below:
caller<func_wrapper_1,func_wrapper_2>::func(0);
caller<func_wrapper_1,func_wrapper_2>::func(1);
I know that every things depends on compiler and compiling option, but lets assume compiler accepts to make these functions inline.
Whether or not a compiler will be smart enough to inline a given situation is up for grabs but I think it may be possible by creating Callable Types by overloading the function call operator.
Something like this:
template<typename Func1, typename Func2>
class caller{
public:
static void func(int v){
if(v) {
// Func1() - creates an object of type Func1
// that object is 'called' using the '()' operator
Func1()();
}else{
Func2()();
}
}
};
struct CallableType1
{
// overloading the function call operator makes objects of
// this type callable
void operator()() const { std::cout << "callable 1" << '\n'; }
};
struct CallableType2
{
void operator()() const { std::cout << "callable 2" << '\n'; }
};
int main()
{
caller<CallableType1, CallableType2> cc;
cc.func(2);
}

C++ store variadic pack from variadic member function

I am working on building a simple 2D game engine and have a class which is capable of acting as a button in a game. I am trying to make a member function such that the user can give the button object a function to run when it is clicked, as well as the parameters for that function. Here is a sample of the code which partially completes the task:
class Entity{
public:
template <typename ... fnParams, typename ... fnArgs>
void Set_Click_Action(void(*action)(fnParams ... params), fnArgs ... args) {
action(args ...); //this works correctly. However, I do not
//want to call the function here, I want
//instead to store the function in the Entity
//object
}
void function(int i, double j, std::string k){
std::cout << i << '\n' << j << '\n' << k << '\n';
}
int main(){
Entity entity;
entity.Set_Click_Action(function, 12, 12.5, std::string("String"));
}
This code works correctly, but does not accomplish my goal. I want to store the function for later use (when the object is clicked). Is it possible to get the function pointer and the variadic pack holding the arguments into member variables? If so, what would be a strategy to do so? I should point out that I am pretty new to using variadic templates in C++. Thanks in advance for any help.
This code works correctly, but does not accomplish my goal. I want to store the function for later use (when the object is clicked). Is it possible to get the function pointer and the variadic pack holding the arguments into member variables? If so, what would be a strategy to do so?
You can save the call of action(), and the args..., in a lambda function and save the lambda function in a std::function
By example
struct Entity
{
std::function<void()> l;
template <typename ... fnParams, typename ... fnArgs>
void Set_Click_Action (void(*action)(fnParams ... params),
fnArgs ... args)
{ l = [=]{ action(args...); }; }
};
Obviously you can add a method Call_Action() to call the function.
The following is a full working example with a Call_Action() method
#include <iostream>
#include <functional>
class Entity
{
private:
std::function<void()> l;
public:
template <typename ... fnParams, typename ... fnArgs>
void Set_Click_Action (void(*action)(fnParams ... params),
fnArgs ... args)
{ l = [=]{ action(args...); }; }
void Call_Action () const
{ l(); }
};
void function (int i, double j, std::string k)
{ std::cout << i << '\n' << j << '\n' << k << '\n'; }
int main()
{
Entity entity;
entity.Set_Click_Action(function, 12, 12.5, std::string("String"));
std::cout << "--- post set\n";
entity.Call_Action();
}

How can I call every instantiated function of a template function?

Is it somehow possible to call every instantiation of a template function without knowing what will get instantiated at write-code-time?
#include <iostream>
template<typename T>
void print_size_of()
{
std::cout << sizeof(T) << "\n";
}
int main()
{
print_size_of<int>();
print_size_of<double>();
//won't work but maybe it shows what i want to do:
template<typename T>
print_size_of<T>();
//is there a syntax so that the compiler replaces that with `print_size_of<int>(); print_size_of<double>();`
}
This is possible; you need to add some static variable inside the function template body to record those instantiations.
In the code below, every instantiated function will have a static variable, whose constructor will register the function pointer to a global registration center:
std::vector<void(*)()> funcs;
struct helper {
explicit helper(void (*f)()) { funcs.push_back(f); }
};
template<typename T>
void print_size_of()
{
static helper _(&print_size_of<T>);
std::cout << sizeof(T) << "\n";
}
int main()
{
print_size_of<int>();
print_size_of<double>();
std::cout << "All instantiation:\n";
for ( auto f : funcs ) {
f();
}
}
EDIT:
This is not strictly recording instantiation. It only records those being called before. If you instantiate it by other methods like taking its address:
void (*f)() = &print_size_of<short>;
but don't call it, then this function pointer won't be registered.
EDIT2:
In fact, it is possible to faithfully record all instantiations. The key point is to associate the instantiation of the function template to the instantiation of a class template. Then a static member of that class will be guaranteed to initialize before entering main() function.
// this one is to make sure `funcs` is initialized
auto &get_funcs() {
static std::vector<void(*)()> funcs;
return funcs;
}
template<void (*f)()>
struct helper {
helper() { get_funcs().push_back(f); }
// this is a static class member, whose initialization is before main()
static helper _ins;
};
template<void (*f)()> helper<f> helper<f>::_ins;
template<typename T>
void print_size_of()
{
// force instantiation of a class
const void *_ = &helper<&print_size_of<T>>::_ins;
std::cout << sizeof(T) << "\n";
}
int main()
{
print_size_of<int>();
print_size_of<double>();
void (*f)() = &print_size_of<short>;
std::cout << "All instantiation:\n";
for ( auto f : get_funcs() ) {
f();
}
}
No, that is not possible.
You can get close by calling every instantiation which was already called once (Use a static variable to register on first call), but that's the best you can do.

c++ functor and function templates

consider this simple and pointless code.
#include <iostream>
struct A {
template<int N>
void test() {
std::cout << N << std::endl;
}
};
int main() {
A a;
a.test<1>();
}
It is a very simple example of a function template. What if however, I wanted to replace A::test with an overloaded operator() to make it a functor?
#include <iostream>
struct A {
template<int N>
void operator()() {
std::cout << N << std::endl;
}
};
int main() {
A a;
a<1>(); // <-- error, how do I do this?
}
Certainly if the operator() took parameters which were dependent on the template, the compiler could possibly deduce the template. But I just can't figure out the proper syntax to specify template parameters with a parameterless functor.
Is there a proper way to do this?
Obviously, this code would work since it bypasses the functor syntax:
a.operator()<1>();
but that kinda defeats the purpose of it being a functor :-P.
You can only call
a.operator()<1>();
but that would not be using a functor. Functors need a non template operator(), as they must be able to be called as varname() and that won't work with your code.
To make it a real functor change your code a template class (functors are classes):
#include <iostream>
template<int N>
struct A {
void operator()() {
std::cout << N << std::endl;
}
};
int main() {
A<1> a;
a();
}
There's not another "direct" way I know other than the:
a.operator()<1>();
syntax. If you're open to changing the code, moving the template parameter to the class would work, or using a (boost|tr1)::bind to make a (boost|tr1)::function object.
You are trying to pass a template parameter to an instance of an object, which as far as I know is not allowed. You can only pass templates parameters to template functions or template objects.
a.test<1>(); and a.operator()<1>(); work because they are serving as template functions.
Use boost::bind (check out boost libraries) to fix it though.
struct A {
void operator()(int n) {
std::cout << n << std::endl;
}
};
int main(int argc, char* argv[]) {
A a;
boost::function<void()> f = boost::bind<void>(a, 1);
f(); // prints 1
return 0;
}
And you don't even have to mess with templates!
You're stuck. Have you considered something like
struct A {
template<int N>
struct B
{
void operator()()
{ std::cout << N << std::endl; }
};
template<int N>
B<N> functor() {return B<N>();}
};
int main()
{
A a;
a.functor<1>()();
}
Nope, there's no way around it. Like you said, you have to either call the operator explicitly (which defeats the purpose), or the template arguments must be able to be deduced by the compiler.