'&' : illegal operation on bound member function expression

Question

This works when I try from a single cpp file with a main function,

 sprintf(smem_options ,
     "#transcode{vcodec=RV24}:smem{"
     "video-prerender-callback=%lld,"
     "no-time-sync},"
    , (long long int)(intptr_t)(void*)&cbVideoPrerender
);

How do I pass function arguments to sprintf within a class?

sprintf(smem_options ,
     "#transcode{vcodec=RV24}:smem{"
     "video-prerender-callback=%lld,"
     "no-time-sync},"
    , (long long int)(intptr_t)(void*)&cbVideoPrerender
);

The error message I get is: error C2276: '&' : illegal operation on bound member function expression

Assuming `cbVideoPrerenderer` is a member function in the second example, you need to say `&Foo::cbVideoPrerenderer`. — Jonathan Wakely, Jan 20 '15 at 16:27
For printing pointers you should use the `"%p"` format, and a `void*` argument. — Some programmer dude, Jan 20 '15 at 16:28
@JoachimPileborg Which will only result in undefined behavior. — James Kanze, Jan 20 '15 at 16:33
If this is just for debug/logging purposes, you could try the follow *hack*. Create an array that is sizeof(p) and memcpy the pointer into it. And then combine that into an integer or other printable value. — Neil Kirk, Jan 20 '15 at 16:42

score 24 · Answer 1 · answered Jan 20 '15 at 16:29

24

Assuming cbVideoPrerenderer is a member function in the second example, you need to say &Foo::cbVideoPrerenderer where Foo is the class it is a member of.

But that will only be valid if it is a static member function. Non-static member functions are not like normal functions, and when you form a pointer-to-member-function with the &Foo::bar syntax the thing you get back cannot be converted to a void* (it is typically something twice as large as a pointer, as it contains information about the object type).

answered Jan 20 '15 at 16:29

Jonathan Wakely

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You are right, it's a non static member function, is there a way I can pass a non static member function as an argument to sprintf? – Jerry Jan 20 '15 at 16:32
@πάνταῥεῖ I think he wants to know how to do it safely. – Neil Kirk Jan 20 '15 at 16:39

score 5 · Answer 2 · answered Jan 20 '15 at 16:43

What you're trying to do is conditionally supported behavior in C++11, and illegal in earlier versions, in both cases. You can't reliably convert a pointer to a function (member or otherwise) to a void*. (I've worked on systems where a pointer to a function was 32 bits, but a void* only 16.)

In practice, most compilers will (illegally, in pre-C++11) ignore the error for non-member functions. Posix requires that function pointers and data pointers be compatible, and they are under Windows as well. (Today: only of the systems where they weren't for me was an early Unix.) As for pointers to members: a pointer to a static member has a type compatible to a pointer to a function (and so will work in practice, if the compiler allows it), but a pointer to a non-static member has a completely different type, usually with a different size, and a different representation. About the only way you can reliably output one is as a series of byte values: put the address in a properly typed variable, take the address of that variable, convert it to unsigned char const*, then use "%02x" to output each byte.

But the real question is why you want to do this. There is nothing that you can reliably do with the value you output, regardlessly of how you output it.

Well from the question (and if I interpret the contents of the string correctly) it looks like the purpose is to pass the function pointer address – via an option *string* – to VLC as a callback for a transcode function. And that simply won’t work with non-static member functions. — Konrad Rudolph, Jan 20 '15 at 16:49
@KonradRudolph That's really the motivation behind the question, That and the fact that there's no guarantee that converting the pointer to a string as he's doing will result in anything useful. — James Kanze, Jan 20 '15 at 16:57

'&' : illegal operation on bound member function expression

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