How to cast a pointer in C++

Question

void foo(void **Pointer);

int main ()
{
    int *IntPtr;

    foo(&((void*)IntPtr));
}

Why do I get an error?

error: lvalue required as unary ‘&’ operand

Thanks

Answer 1

void foo(void **Pointer);

int main ()
{
    int *IntPtr;

    foo((void**)&IntPtr);
}

Answer 2

When you do

(void*)IntPtr

you create a temporary variable, which is only an rvalue, and so can't be dereferenced.

What you need to do is:

int main()
{
  int* IntPtr;
  void* VoidPtr = (void*)IntPtr;
  foo(&VoidPtr);
}

or equivalent

Answer 3

(void*) is not an lvalue, it is kind of a casting operator, you need to have the ampersand to the immediate left of the variable (lvalue). This should be right:

foo(((void**)&IntPtr));

Answer 4

More C++ style:

foo( reinterpret_cast< void** >( IntPtr ) );

But remember, according to Standard such a cast is implementation specific. Standard gives no guarantees about behavior of such cast.

Answer 5

As others point out, you have the order of the cast and the & wrong. But why do you use void** at all? That means that you accept a pointer to a void pointer. But that's not at all what you want. Just make the parameter a void*, and it will accept any pointer to some object:

void foo(void*);

int main () {
    int *IntPtr;
    foo(&IntPtr);
    assert(IntPtr == NULL);
}

That's what void* is for. Later, cast it back using static_cast. That's a quite restrictive cast, that doesn't allow dangerous variants, unlike the C style cast (type):

void foo(void* p) {
    int** pint = static_cast<int**>(p);
    *pint = NULL;
}

If the function takes pointers to void*, then that function can't accept pointers to int*. But if the function accepts either or, then the function should accept void*, and you should cast to the proper type inside the function. Maybe paste what you really want to do, we can help you better then. C++ has some good tools available, including templates and overloading, both of which sound helpful in this case.