C++ Programming :: function pointer to extern?


	function pointer to extern? I'm relatively new to externs and function pointers but have inherited the task of modifying some existing code. I have something like this. extern void myExternFunc (MenuItem item, void param) void (fptr)(MenuItem item, void * param) = myExternFunc; This doesn't compile. I don't understand why. Obviously something like this: extern void myFunc (MenuItem item, void param) { } void (fptr)(MenuItem item, void * param) = myFunc compiles fine. The only thing I can think to check is that you can creat function pointer to externs. Thanks. Travis wrote: > I'm relatively new to externs and function pointers but have inherited > the task of modifying some existing code. I have something like this. > extern void myExternFunc (MenuItem item, void param) A semicolon is missing at the end of the previous line. > void (fptr)(MenuItem item, void * param) = myExternFunc; > This doesn't compile. I don't understand why. Obviously something like > this: > extern void myFunc (MenuItem item, void param) { } > void (fptr)(MenuItem item, void * param) = myFunc > compiles fine. The only thing I can think to check is that you can > creat function pointer to externs. All functions are 'extern' by default, IIRC, unless they are 'static'. V -- Please remove capital 'A's when replying by e-mail I do not respond to top-posted replies, please don't ask Travis wrote: > I'm relatively new to externs and function pointers but have inherited > the task of modifying some existing code. I have something like this. > extern void myExternFunc (MenuItem item, void param) > void (fptr)(MenuItem item, void * param) = myExternFunc; > This doesn't compile. What does "doesn't compile" mean? You should post a minimal, but complete program together with the compiler error message. Travis wrote: > I'm relatively new to externs and function pointers but have inherited > the task of modifying some existing code. I have something like this. > extern void myExternFunc (MenuItem item, void param) > void (fptr)(MenuItem item, void * param) = myExternFunc; > This doesn't compile. I don't understand why. Obviously something like > this: As Victor already mentioned, the keyword "extern" is unneeded here. Beyond that, all you need is a ';' after the declaration of myExternFunc. Or you can provide a definition as below (and if you don't, you'll have to define it /somewhere/). Mark - Hide quoted text - > extern void myFunc (MenuItem item, void param) { } > void (fptr)(MenuItem item, void * param) = myFunc > compiles fine. The only thing I can think to check is that you can > creat function pointer to externs. > Thanks. Forgive me. The ';' is there, I just didn't select everything before doing a copy/paste. Second, the extern is there because the function is actually declared / implemented somewhere else. The extern code is the part I inherited. The line below it with the function pointer is what I'm adding. I have made some progress and I think a better question might be asking if the following is allowed. - a struct containing a few char * and a function pointer - a linked list w/ the struct as nodes - the head of the linked list passed as a void pointer to elsewhere in the app Something I'm doing is freezing up my machine entirely. I have a feeling it's dereferencing related. Travis wrote: > Forgive me. The ';' is there, I just didn't select everything before > doing a copy/paste. Please quote context when replying. Like I've done here. > Second, the extern is there because the function is actually > declared / implemented somewhere else. The extern code is the part I > inherited. The line below it with the function pointer is what I'm > adding. You're missing the point. The "extern" is not required. Without additional specifiers, functions have external linkage by default. The "extern" is entirely superfluous. > I have made some progress and I think a better question might be > asking if the following is allowed. > - a struct containing a few char * and a function pointer > - a linked list w/ the struct as nodes > - the head of the linked list passed as a void pointer to elsewhere in > the app This is a very vague question which makes it hard to give you a useful answer. From what little you've specified, it could be fine, but without seeing what you're doing, no one really knows. - Hide quoted text - > Something I'm doing is freezing up my machine entirely. I have a > feeling it's dereferencing related. See my insetrs bellow: Travis wrote: > I'm relatively new to externs and function pointers but have inherited > the task of modifying some existing code. I have something like this. > extern void myExternFunc (MenuItem item, void param) semicolon ";" is missing after above line. > void (fptr)(MenuItem item, void * param) = myExternFunc; > This doesn't compile. I don't understand why. Obviously something like > this: > extern void myFunc (MenuItem item, void param) { } when doing extern, you should not declare the body. - Hide quoted text - > void (fptr)(MenuItem item, void * param) = myFunc > compiles fine. The only thing I can think to check is that you can > creat function pointer to externs. > Thanks. On Jun 4, 9:26 pm, Haro Panosyan <h@ti.com> wrote: [...] > > extern void myFunc (MenuItem item, void param) { } > when doing extern, you should not declare the body. That's a frequent convention, but is not based on anything in the language. In a function declaration, "extern" has no effect as to whether the declaration is a definition or not---all it says is that the linkage is external, which is the default for functions anyway. As a matter of style, I do put the extern in front of the declaration, in the header, and I don't put it in front of the definition, but it is strictly a matter of style; the opposite is also fine, as far as the compiler is concerned. -- James Kanze (GABI Software) email:james.ka@gmail.com Conseils en informatique oriente objet/ Beratung in objektorientierter Datenverarbeitung 9 place Smard, 78210 St.-Cyr-l'cole, France, +33 (0)1 30 23 00 34