In void pointr in c, why there is a '*' after 'int' ?

Question

printf("void ptr points to %d\n",*((int *)ptr));

Answer 1

A void pointer is a pointer to some data with no type attached. We are basically saying to the compiler "this pointer points to 'some' data, but I only care about the memory location of the data, nothing else".

So to access a void pointer, you need to 'cast' it to a pointer of some data type. In the code above

(int*)ptr

casts 'ptr' to a 'int pointer'. Now we can dereference (*) it to get the int stored in memory pointed by 'ptr'

*((int*)ptr)

It's basically doing
*ptr
but after casting ptr to an int pointer, hence
*((int*)ptr)

Answer 2

Dmitry
Could you expand on the line
"(-46864457) which, as an address, does not exist at all, since the address is read as a hexadecimal number, and not as an integer."
?

What do mean by "the address is read as a hexadecimal number, and not as an integer"? I mean, hexadecimal numbers are.... integers right? Pointers are displayed in hexadecimal format, but they are numbers. You are right that the address (-46864457) doesn't exist, but that's because you are casting a pointer to a signed int. Try casting a pointer to (unsigned long) and printing it, that will give you the same number in decimal format as you would get by printing it as a pointer