assign func(const void ) to func(void ) in C

Question

I have a function that takes a function pointer and apply it over a list. It's prototype is as follows:

void applyList(List *, void applyFunc(void *));

Now I want to supply a function that prints the element in the list. Since I'm not modifying the list element, the print function looks like this:

void printNode(const void *nodeData){
    Data *dPtr=(Data*)nodeData;
    printf("%s", dPtr->str );
}

//Usage
applyList(list, printNode);

However I got an compiler error saying

expected ‘void (*)(void *)’ but argument is of type ‘void (*)(const void *)’

I don't want to add const to apply's prototype, because I might supply some function that modifies the data. Any suggestion on how to deal with this situation?

score 3 · Accepted Answer · answered Sep 29 '11 at 18:43

3

Unfortunately, in C function pointers are not polymorphic. At all. Attempting to force a cast leads to undefined behavior.

Your only standard-conforming way out is to write a wrapper function:

static void wrap_printNode(void *nodeData) {
  printNode(nodeData);
}

Note, however, that your const is kind of useless here:

Data *dPtr=(Data*)nodeData;

You just casted away the const declaration. Try to keep it on there:

const Data *dPtr=(const Data*)nodeData;

answered Sep 29 '11 at 18:43

bdonlan

224,562
31
268
324

Technically there is a very small and verbose degree of polymorphism (http://stackoverflow.com/a/559671/163349), but you are correct that not even `((void(*)(void*))void_from_void_const_ptr)(p)` is defined. – David X Dec 20 '14 at 21:08

score 0 · Answer 2 · answered Sep 29 '11 at 18:43

0

I don't want to add const to apply's prototype, because I might supply some function that modifies the data.

Then there is no sane solution. The data isn't yours to modify, which is why it's marked const.

answered Sep 29 '11 at 18:43

Ignacio Vazquez-Abrams

776,304
153
1,341
1,358

assign func(const void *) to func(void *) in C

2 Answers2

assign func(const void ) to func(void ) in C