c++ - How are function pointers type unsafe

Question

First of all type-safe means that anything that a compiler can catch straight away if done incorrectly.

Now, I heard function pointers are not type safe however whenever I tried to use them incorrectly the compiler did report errors for me. So, how is it type unsafe ?

E.g This is a function prototype that takes in a function pointer

void SortElements(void* MyArray, unsigned int iNumofElems,size_t size, int(*compare_funct)(void* First,void* SecondElem))

I have defined few functions to pass to it as:

int MySortAsc(void* First, void* Second);
void MyFunct2();
void MyFunct3(void* First);

The code only compiles for:

SortElements(MyArray, 10, sizeof(DataType), &MySortAsc); //Compiles
SortElements(MyArray, 10, sizeof(DataType), &MyFunct2);  //Fails

Any idea how can I mis-use function pointers here ?

Is it because of this:

void (*functionPointer)();
...
int integer = 0xFFFFFFFF;     
functionPointer = (void(*)())integer;      
functionPointer();

Answer: What I got to see is that function pointers in C++ are type safe. Ofcourse, they can be used in an unsafe manner by casting it incorectly but that does not make them a reason to be called as type unsafe. .NET delegates are strongly typed as well and to me it looks like both are type safe.

Answer 1

So, how is it type unsafe ?

void SortElements(void* MyArray,              // what type is pointed here?
                  unsigned int N,             // Are there really N elements?
                  size_t size,                // Is the size correct?
                  int(*cmp)(void*,void*));    // Is this the correct function?

The code that you present is type-unsafe, not because of the function pointer but rather because of the use of void* in both the SortElements signature and the signature of the function pointer.

The reason why this is unsafe is because the caller has the whole responsibility of passing the right arguments, and the compiler cannot ensure that the pointer MyArray points to a contiguous memory region that holds iNumofElems each of which has the size offered in the interface. If the programmer makes a mistake, the compiler will not be able to help there, if a maintainer modifies the type stored in the array (size changes) or the number of elements, the compiler will not be able to detect it and tell you that you need to update the call to SortElements. Finally, because the function pointer that is passed also uses void*, the signature of a comparator that compares apples and pears is exactly the same, and the compiler cannot help if you pass the incorrect function pointer.

struct Apple {
   int weight;
};
struct Pear {
   double weight;
};
int compare_pears( void * pear1, void * pear2 ) {
   return static_cast<Pear*>(pear1)->weight - static_cast<Pear*>(pear2)->weight;
}
int main() {
   Apple apples[10];
   SortElements( apples, 20, sizeof(Pear), compare_pears );
}

While the compiler is able to verify that the signature of the function pointer matches the signature that the function needs, the function pointer itself is unsafe, and allows you to pass a comparator for basically anything.

Compare that with this other alternative:

template <typename T, std::size_t N>
void SortElements( T (&array)[N], int (*cmp)( T const &, T const & ) );

Here the compiler will infer the type of the elements T and the size of the array N from the call. There is no need to pass the size of T, as the compiler knows it. The comparator function passed to this version of SortElements is strongly typed: it takes two constant references to the type of the element stored in the array and returns an int. If we tried this in the previous program:

int compare_pears( Pear const & lhs, Pear const & rhs );
int compare_apples( Apple const & l, Apple const & r );
Apple array[10];
//SortElements( array, compare_pears );   // Error!!!!
SortElements( array, compare_apples );    // Good!

You cannot mistake the size of the array or the size of the elements, if someone changes the type Apple, the compiler will pick it up, if the size of the array changes, the compiler will pick it up. You cannot mistake the comparator that is passed to the function as the compiler will also pick it up. Now the program is type safe, even if it uses function pointers (that might have an impact in performance as they inhibit inlining, which is why std::sort is usually faster than qsort)