Explanation of converting from void* to int in c

Question

in my code i have:

int number_compare(void *val1, void *val2) {
    if (*(int*) val1 < *(int*s) val2) {

So basically to convert from void* to int i have to cast *(int *).

This works and give me the correct results, could someone please tell me why though or point me to a thread that explains it. I have already looked and cannot find an answer I understand.

Answer 1

That's not converting a void * to an int. It's interpreting whatever the void * is pointing at as an int. Break it down:

val1           // void pointer - not dereferenceable
(int *)val1    // pointer to 'int'
*(int *)val1   // the 'int' being pointed to

So your function is getting passed two pointers: it then interprets them as pointers to int and dereferences them, comparing the two int values being pointed to.

In contrast, converting from a void * to an int would look something like this:

int x = (int)val1;

But that's almost certainly not what you want - first because int is signed, and pointers aren't, and second because int and pointer types might not be the same size.

Answer 2

first thing void pointer cannot be defreferenced. maybe because it doesn't yet know how to fetch data. (i.e) if its char should fetch 1 byte, int 4 bytes...

so here first you are converting some address (void pointer) to int pointer.

(int*) val1;

later to fetch the value from that address [now the system knows it should take data from 4 bytes from that address : val1].

*(int*)val1 

this 'll give you the data in that address.

This is formally said to be casting "(data_type) data" casting the data to specified data_type;