Dereference void pointer

Question

Even after casting a void pointer, I am getting compilation error while dereferencing it. Could anyone please let me know the reason of this.

int lVNum = 2;
void *lVptr;
lVptr = (int*)&lVNum;

printf("\nlVptr[60 ] is  %d \n",lVptr[1]);

Answer 1

It doesn't make sense to dereference a void pointer. How will the compiler interpret the memory that the pointer is pointing to? You need to cast the pointer to a proper type first:

int x = *(int*)lVptr;

Answer 2

printf("\nlVptr[60 ] is %d \n", *(int*)lVptr);

This will cast the void pointer to a pointer to an int and then dereference it correctly.

If you want to treat it as an array (of one), you could do a slightly ugly ((int *)lVptr)[0]. Using [1] is out of bounds, and therefore not a good idea (as for lVptr[60]...)

Answer 3

It's still a void* because that's what you declared it as. Any pointer may be implicitly converted to a void*, so that cast does nothing and you are left with a pointer to void just as you began with.

You'll need to declare it as an int*.

void *some_ptr = /* whatever */;
int *p = (int*)some_ptr;
// now you have a pointer to int cast from a pointer to void

Note that the cast to an int* is also unnecessary, for the same reason you don't have to (and should not) cast the return value of malloc in C.

void*'s can be implicitly converted to and from any other pointer type. I added the cast here only for clarity, in your code you would simply write;

int *p = some_void_ptr;

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Also, this:

lVptr[1]

Is wrong. You have a pointer to a single int, not two. That dereference causes undefined behavior.