invalid cast from type âvoid*â to class member_function pointer

Question

typedef A::T*(Processor::*myMethodType)(const FDS::T*);

myMethodType temp = NULL;

temp = reinterpret_cast<myMethodType> (myInterface.findMap(moname)))

Note: Processor is a class which instantiated Template class . myMethodType is the member-pointer-func pointer .

Note : findMap is returning a void* as seen below and myinterface is a map of

template <class Implementor>

void* Interface<Implementor>::findMap(std::string &name){
    if (myInterface.find(moname) != myInterface.end()) {
        return myInterface.find(moname)->second;
    }

    return NULL;
}

========================================================

Getting below error -

error: invalid cast from type âvoid*â to type âProcessor::myMethodType {aka T* (Processor::*)(const T*)}â
      if((temp = reinterpret_cast<myMethodType> (myInterface.findMap(moname))) != NULL)

Question - why am I getting this error, though I am casting it to member_class_pointer? why Invalid conversion?

Tried till now :

Tried below approach from --- https://stackoverflow.com/questions/1307278/casting-between-void-and-a-pointer-to-member-function

void myclass::static_fun(myclass *instance, int arg)
{
   instance->nonstatic_fun(arg);
}

but no luck so far !!

`is returning a void*` - right, so it's not returning `myMethodType` nor `FDS::T*(Processor::*myMethodType)(const FDS::T*);`, it's returning `void*`. What is your question? — KamilCuk, Sep 11 '19 at 07:53
You first should think of C++ programming in general. Having a void* in your code should always alarm you! — Klaus, Sep 11 '19 at 08:14

VLL · Answer 1 · 2019-09-11T08:07:55.740

0

You cannot cast between function pointers and object pointers (such as void*). If you want a map with multiple function pointer types, you can convert them to some more function pointer type, such as void*(*)(). Function pointers must be converted back to the original type before calling them.

If you don't need multiple function pointer types, you can use a map of myMethodType.

edited Sep 11 '19 at 08:07

answered Sep 11 '19 at 08:00

VLL

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you mean to say - I should have use a map of `void*(*)()` instead of `void *` ? – dontask Sep 11 '19 at 08:05
1

Why not use a map of `myMethodType` ? – KamilCuk Sep 11 '19 at 08:07
as myMethodType is a member-function -pointer to a template class , which is instantiated by different `types` . – dontask Sep 11 '19 at 08:13

score 0 · Answer 2 · answered Sep 11 '19 at 08:05

0

A regular pointer is different from pointer-to-member. Since reinterpret_cast conversion (see https://en.cppreference.com/w/cpp/language/reinterpret_cast) a pointer can be converted to another pointer or to an integral type, but not to pointer-to-member

answered Sep 11 '19 at 08:05

P. Dmitry

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but void pointer does not know which type it is pointing to ? so it can not be considered as a regular pointer, it may be a pointer to member, please correct me. – dontask Sep 11 '19 at 08:11
Pointer and pointer-on-member it's just different types. You can not cast it to each other. See https://en.cppreference.com/w/cpp/language/pointer `Typically, mentions of "pointers" without elaboration do not include pointers to (non-static) members.` – P. Dmitry Sep 11 '19 at 08:16
Probably you should change return type on `void*` for `myMethodType` and return `myMethodType` from `find` – P. Dmitry Sep 11 '19 at 08:21
We cant change to myMethodType , as this is a template class – dontask Sep 12 '19 at 08:30
Do you implement this class? If so, you can freely change the return type of methods to `void*` – P. Dmitry Sep 12 '19 at 08:38

invalid cast from type âvoid*â to class member_function pointer

2 Answers2