A void* does not mean anything. It is a pointer, but the type that it points to is not known.

It's not that it can return "anything". A function that returns a void* generally is doing one of the following:

• It is dealing in unformatted memory. This is what operator new and malloc return: a pointer to a block of memory of a certain size. Since the memory does not have a type (because it does not have a properly constructed object in it yet), it is typeless. IE: void.
• It is an opaque handle; it references a created object without naming a specific type. Code that does this is generally poorly formed, since this is better done by forward declaring a struct/class and simply not providing a public definition for it. Because then, at least it has a real type.
• It returns a pointer to storage that contains an object of a known type. However, that API is used to deal with objects of a wide variety of types, so the exact type that a particular call returns cannot be known at compile time. Therefore, there will be some documentation explaining when it stores which kinds of objects, and therefore which type you can safely cast it to.

This construct is nothing like dynamic or object in C#. Those tools actually know what the original type is; void* does not. This makes it far more dangerous than any of those, because it is very easy to get it wrong, and there's no way to ask if a particular usage is the right one.

And on a personal note, if you see code that uses void*'s "often", you should rethink what code you're looking at. void* usage, especially in C++, should be rare, used primary for dealing in raw memory.

Is it possible to dereference the void pointer without type-casting in C programming language...

No, void indicates the absence of type, it is not something you can dereference or assign to.

is there is any way of generalizing a function which can receive pointer and store it in void pointer and by using that void pointer we can make a generalized function..

You cannot just dereference it in a portable way, as it may not be properly aligned. It may be an issue on some architectures like ARM, where pointer to a data type must be aligned at boundary of the size of data type (e.g. pointer to 32-bit integer must be aligned at 4-byte boundary to be dereferenced).

For example, reading uint16_t from void*:

/* may receive wrong value if ptr is not 2-byte aligned */
uint16_t value = *(uint16_t*)ptr;
/* portable way of reading a little-endian value */
uint16_t value = *(uint8_t*)ptr
                | ((*((uint8_t*)ptr+1))<<8);

Also, is pointer arithmetic with void pointers possible...

Pointer arithmetic is not possible on pointers of void due to lack of concrete value underneath the pointer and hence the size.

void* p = ...
void *p2 = p + 1; /* what exactly is the size of void?? */

A pointer to void is a "generic" pointer type. A void * can be converted to any other pointer type without an explicit cast. You cannot dereference a void * or do pointer arithmetic with it; you must convert it to a pointer to a complete data type first.

void * is often used in places where you need to be able to work with different pointer types in the same code. One commonly cited example is the library function qsort:

void qsort(void *base, size_t nmemb, size_t size, 
           int (*compar)(const void *, const void *));

base is the address of an array, nmemb is the number of elements in the array, size is the size of each element, and compar is a pointer to a function that compares two elements of the array. It gets called like so:

int iArr[10];
double dArr[30];
long lArr[50];
...
qsort(iArr, sizeof iArr/sizeof iArr[0], sizeof iArr[0], compareInt);
qsort(dArr, sizeof dArr/sizeof dArr[0], sizeof dArr[0], compareDouble);
qsort(lArr, sizeof lArr/sizeof lArr[0], sizeof lArr[0], compareLong);

The array expressions iArr, dArr, and lArr are implicitly converted from array types to pointer types in the function call, and each is implicitly converted from "pointer to int/double/long" to "pointer to void".

The comparison functions would look something like:

int compareInt(const void *lhs, const void *rhs)
{
  const int *x = lhs;  // convert void * to int * by assignment
  const int *y = rhs;

  if (*x > *y) return 1;
  if (*x == *y) return 0;
  return -1;
}

By accepting void *, qsort can work with arrays of any type.

The disadvantage of using void * is that you throw type safety out the window and into oncoming traffic. There's nothing to protect you from using the wrong comparison routine:

qsort(dArr, sizeof dArr/sizeof dArr[0], sizeof dArr[0], compareInt);

compareInt is expecting its arguments to be pointing to ints, but is actually working with doubles. There's no way to catch this problem at compile time; you'll just wind up with a missorted array.