Function pointers in C

Pascal Cuoq - 24th Aug 2013

This post contains a complete list of everything a C program can do with a function pointer, for a rather reasonable definition of “do”. Examples of things not to do with a function pointer are also provided. That list, in contrast, is in no way exhaustive.

What a C program can do with a function pointer

Convert it to a different function pointer type

A function pointer can be converted to a different function pointer type. The C99 standard's clause starts:

“A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer.”

Call the pointed function with the original type

Clause continues:

“If a converted pointer is used to call a function whose type is not compatible with the pointed-to type, the behavior is undefined.”

Alright, so the above title is slightly sensationalistic: the pointed function can be called with a compatible type. After typedef int t;, the types t and int are compatible, and so are t (*)(t) and int (*)(int), the types of functions taking a t and returning a t and of functions taking an int and returning an int, respectively.

There is no third thing a C program can do with a function pointer

Seriously. The C99 standard has uintptr_t, a recommended integer type to convert data pointers to, but there is not even an equivalent integer type to store function pointers.

What a C program cannot do with a function pointer

Convert it to an ordinary pointer

Function pointers should not be converted to char * or void *, both of which are intended for pointers to data (“objects” in the vocabulary of the C standard). Historically, there has been plenty of reasons why pointers to functions and pointers to data might not have the same representation. With 64-bit architectures, the same reasons continue to apply nowadays.

Call the pointed function with an incompatible type

Even if you know that type float is 32-bit, the same as int on your platform, the following is undefined:

void f(int x); 
int main(){ 
  void (*p)(float) = f; 

The line void (*p)(float) = f;, which defines a variable p of type “pointer to function that takes a float”, and initializes it with the conversion of f, is legal as per However, the following statement, (*p)(3); is actually equivalent to (*p)((float)3);, because the type of p is used to decide how to convert the argument prior to the call, and it is undefined because p points to a function that requires an int as argument.

Even if you know that the types int and long are both 32-bit and virtually indistinguishable on your platform (you may be using an ILP32 or an IL32P64 platform), the types int and long are not compatible. Josh Haberman has written a nice essay on this precise topic.

Consider the program:

void f(int x); 
int main(){ 
  void (*p)(long) = f; 

This time the statement is equivalent to (*p)((long)3); and it is undefined even if long and int are both 32-bit (substitute long and long long if you have a typical I32LP64 platform).

Lastly the example that prompted this post was in a bit of Open-Source code the creation of a new execution thread. The example can be simplified into:

void apply(void (*f)(void*)  void *arg) 
void fun(int *x){ 
  // work work 
  *x = 1; 
int data; 
int main(){ 
  apply(fun  &data); 

The undefined behavior is not visible: it takes place inside function apply() which is a standard library function (it was pthread_create() in the original example). But it is there: the function apply() expects a pointer to function that takes a void* and applies it as such. The types int * and void * are not compatible and neither are the types of functions that take these arguments.

Note that gcc -Wall warns about the conversion when passing fun to apply():

t.c:11: warning: passing argument 1 of ‘apply’ from incompatible pointer type 

Fixing this warning with a cast to void (*)(void*) is a programmer mistake. The bug indicated by the warning is that there is a risk that fun() will be applied with the wrong type and this warning is justified here since fun() will be applied with the wrong type inside function apply(). If we “fix” the program this way:

$ tail -3 t.c 
int main(){ 
  apply((void (*)(void*))fun  &data); 
$ gcc -std=c99 -Wall t.c 

The explicit cast to (void (*)(void*) silences the compiler but the bug is still in the same place in function apply().

Fortunately gcc -std=c99 -Wall is not the only static analyzer we can rely on. Frama-C's value analysis warns where the problem really is in function apply() and it warns for both the version with implicit conversion and the version with explicit cast:

$ frama-c -val t.c 
[value] computing for function apply <- main. 
        Called from t.c:14. 
t.c:3:[value] warning: Function pointer and pointed function 'fun'  have incompatible types: 
        void (void *) vs. void (int *x). assert(function type matches) 

The correct way to use function apply() without changing it is to make a function with the correct type for it and to pass that function to apply():

void stub(void *x){ 
  apply(stub  &data); 

Note that in the above x is implicitly converted when passed to function fun() the same way that &data is implicitly converted to void* when passed to apply().


There is almost nothing you can do in C with a function pointer. The feature is still very useful and instills a bit of genericity in an otherwise decidedly low-level language.

Function pointers are not often used in the standard library considering: qsort() is with pthread_create() another of the few functions that requires a function pointer. Like it it is often misused: it has its own entry in the C FAQ.

Jens Gustedt provided advice in the writing of this post.

Pascal Cuoq
24th Aug 2013