Error C2276 Illegal Operation
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Std::invoke No Matching Overloaded Function Found
about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss member function pointer Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Passing a pointer non-standard syntax; use '&' to create a pointer to member to a class member function as a parameter up vote 10 down vote favorite 6 I have written a small program where I am trying to pass a pointer to member function of a class to another function. Can you please help me and where I am going wrong..? #include
Std::bind Member Function
* ptr); void call_cb_func(); }; void test::get_pc(){ cout << "PC" << endl; } void test::set_cb_ptr( void *ptr){ cb_func = (test::callback_func_ptr)ptr; } void test::call_cb_func(){ cb_func(); } int main(){ test t1; t1.set_cb_ptr((void *)(&t1.get_pc)); return 0; } I get the following error when I try to compile it. error C2276: '&' : illegal operation on bound member function expression c++ function-pointers share|improve this question asked Aug 9 '13 at 11:41 ajay bidari 2622417 I always find newty.de/fpt/index.html very useful. –arne Aug 9 '13 at 11:43 2 Member functions aren't functions. The type you need is void (test::*)(void *)... –Kerrek SB Aug 9 '13 at 11:43 1 Kerrek SB is right. However if you intend to call the same member for different classes and instances you should think of inheritance and virtuals... –Manuel del Castillo Aug 9 '13 at 12:04 add a comment| 1 Answer 1 active oldest votes up vote 15 down vote You cannot cast a function pointer to void*. If you want a function pointer to point to a member function you must declare the type as ReturnType (ClassType::*)(ParameterTypes...) Further you cannot declare a function pointer to a bo
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Std::function
each other. Join them; it only takes a minute: Sign up Why does this pointer to C++ function code generate a compile error? up vote 1 down vote favorite Can anyone solve this? I can’t seem to find http://stackoverflow.com/questions/18145874/passing-a-pointer-to-a-class-member-function-as-a-parameter the solution anywhere, but I see no logical reason why the line below (with the comment showing the compile error) should be a problem. Note: This question is a derivative of How can a C++ base class determine at runtime if a method has been overridden? class MyClass { typedef void (MyClass::*MethodPtr)(); virtual void Method() { MethodPtr a = &MyClass::Method; // legal MethodPtr b = &Method; // error C2276: ‘&’ : illegal operation on bound member function http://stackoverflow.com/questions/1802059/why-does-this-pointer-to-c-function-code-generate-a-compile-error expression if (a == b) // this method has not been overridden? throw “Not overridden”; } }; c++ visual-c++ share|improve this question edited Nov 26 '09 at 13:35 MSalters 107k882221 asked Nov 26 '09 at 7:24 AndrewR 3,89382945 add a comment| 1 Answer 1 active oldest votes up vote 12 down vote accepted ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. This takes care of name mangling. So what you are trying to do will not work in a standards compliant C++ compiler. share|improve this answer edited Nov 26 '09 at 8:06 answered Nov 26 '09 at 7:28 Vijay Mathew 19.6k14183 If you could do this, C++ would have something like [closures][1]. [1]: en.wikipedia.org/wiki/Closure_%28computer_science%29 –Conrad Meyer Nov 26 '09 at 7:43 @Conrad Meyer en.wikipedia.org/wiki/Function_object#In_C_and_C.2B.2B –Vijay Mathew Nov 26 '09 at 8:05 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you agree to the privacy policy and terms of service. Not the answer you're looking for? Browse other questions tagged c++ visual-c++ or ask your own question. asked 6 years ago viewed 2400 time
file I have this: void (*ptr)(unsigned char*, int, int); void func01(unsigned char*, int, int); And in the implementation file I have http://www.cplusplus.com/forum/general/102086/ this: ptr=&func01; OR C::ptr=&C::func01; etc. Still get errors And this is the https://www.reddit.com/r/learnprogramming/comments/2gavlf/c11_what_is_the_proper_way_to_pass_a_member/ error I get: error C2440: '=' : cannot convert from 'void (__thiscall C::* )(unsigned char *,int,int)' to 'void (__cdecl *)(unsigned char *,int,int)' 1> There is no context in which this conversion is possible error C2276: '&' : illegal operation on bound member function expression If I use: ptr=func01; I member function get: error C3867: 'C::func01': function call missing argument list; use '&C::func01' to create a pointer to member error C2440: '=' : cannot convert from 'void (__thiscall C::* )(unsigned char *,int,int)' to 'void (__cdecl *)(unsigned char *,int,int)' 1> There is no context in which this conversion is possible SO, if I go ptr=&C::func01; I get this: error C2440: '=' : cannot convert error c2276 illegal from 'void (__thiscall C::* )(unsigned char *,int,int)' to 'void (__cdecl *)(unsigned char *,int,int)' 1> There is no context in which this conversion is possible OR ptr=&func01; I get: error C2276: '&' : illegal operation on bound member function expression Somewhere I'm making some kind of trivial yet fundamental mistake but I can't figure it out and any help is appreciated. RON Last edited on May 14, 2013 at 2:05pm UTC May 14, 2013 at 2:05pm UTC kooth (736) Could you please post the definition for func01()? May 14, 2013 at 2:15pm UTC RonInNewYork (4) void func01(unsigned char*, int, int); ONE THING I FORGOT TO MENTION: func01 is a member function (in class C). And the function pointer is also a member of class C. RON May 14, 2013 at 2:17pm UTC Bourgond Aries (415) Please use code tags. I believe the correct syntax is for member functions is: return_type (ClassName::* pointer_name)(input...); And thus becomes: 1
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