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Passing typedef function pointer as an argument problem


By : user2174593
Date : October 17 2020, 08:10 PM
will be helpful for those in need It is not necessary to give function parameters names in function declarations.
code :
UniqueOrderedList uniqueOrderedListCreate(copyElements, freeElements,
    elementsEquals, elementGreaterThan);
UniqueOrderedList uniqueOrderedListCreate(copyElements a, freeElements b,
    elementsEquals c, elementGreaterThan d)
{
    /* ... */
}


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C : function pointer and typedef problem


By : abhi4u1947
Date : March 29 2020, 07:55 AM
I hope this helps you . Your dummy destructor is declared to return a void pointer, not a void. This declaration does not match the typedef for your destructor function pointer.

How to use typedef to simplify tempate function argument that is a function pointer


By : Christoeffer John Es
Date : March 29 2020, 07:55 AM
hope this fix your issue In C++11, you can get the rough equivalent of a template typedef by using the using keyword. This still allows N to be deduced from the argument:
code :
template <unsigned N>
using fooP = int (*) (const char (*)[N]);

template <unsigned N>
int foo (fooP<N> ptr) {
  return ptr(0);
}

int bar(const char (*p)[2]) {
  return 0;
}

int main() {
  return foo(bar);
}

deduce template argument for typedef'd function pointer


By : user3245599
Date : March 29 2020, 07:55 AM
this will help Not with your current setup. Everything to the left of ::* is a non-deduced context, because a metafunction can apply arbitrary Turing-complete transformations on the type, and in any event there's no guaranteed one-to-one mapping between the original type and the result. So the language doesn't even try.
But an alias template will work:
code :
template <class T>
using FuncPtr = void (*)(const T&);

template <typename T>
void f3(FuncPtr<T> fp) { return f_(fp); }

pre-typedef'ing a variadic-function-pointer argument


By : WobiHH
Date : March 29 2020, 07:55 AM
This might help you I believe this may be related to the following text which I copied from this answer that itself takes from the C++ standard in 14.5.7 [temp.alias] paragraph 2:

Passing a typedef method as pointer function


By : Kelly Diaz
Date : March 29 2020, 07:55 AM
this will help The problem here is that ptr in bind_regex_member points to const T and your method InterpreterClass::MatchAddressCallback is non const. Basicly like this:
code :
InterpreterClass i;
const InterpreterClass* myPtr = &i;
MatchState myMs;
myPtr->MatchAddressCallback("", 0, myMs); // OUCH! myPtr points to const T and MatchAddressCallback is non const member function
class Interpreter
{
public:
    void init();
    GlobalMatchCallback MatchAddressCallback; // <----------- HERE
};
#ifndef _INTERPRETER_h
#define _INTERPRETER_h

#include <Arduino.h>
#include "Regexp.h"

template<class T, void(T::*PTR)(const char *, const unsigned int, const MatchState &), size_t I>
struct bind_regex_member
{
    typedef void(*fn_type)(const char *, const unsigned int, const MatchState &);
    explicit bind_regex_member(T* _ptr)
    {
        ptr = _ptr;
    }
    static void func(const char * match, const unsigned int length, const MatchState & ms)
    {
        (ptr->*PTR)(match, length, ms);
    }
    operator fn_type()
    {
        return &func;
    }
private:
    static T*  ptr;
};

template<class T,

void(T::*PTR)(const char *, const unsigned int, const MatchState &), size_t  I>
T* bind_regex_member<T, PTR, I>::ptr = NULL;

class InterpreterClass
{
 public:
    void init();

    void MatchAddressCallback(const char * match, const unsigned int length, const MatchState & ms);
};

extern InterpreterClass Interpreter;

#endif
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