Converting int to int* then back to int

Question

I am missing something obvious here but consider the following,

int k = 10;
int *p = &k;
int i = (int)p;

above produces,

warning: cast from pointer to integer of different size

and following,

int k = 10;
int *p = &k;
int i = p;

causes,

warning: initialization makes integer from pointer without a cast

a bit of googling lead me to,

Converting a pointer into an integer

which advices using uintptr_t,

int k = 10;
int *p = &k;
int i = (uintptr_t)p;

above works no errors no warnings. So my question is k is an int and p is an int pointer pointing to k why do i get an error dereference p?

Answer 1

int k = 10;
int *p = &k;
int i = *p;

Note the * before p on the last line. Dereference the pointer to get the int it points to.

EDIT: I'm not very familar with uintptr_t, but I ran the code int i = (uintptr_t)p; The result that is stored into i is the memory address of k, not 10. The cast tells the compiler that you really want to turn the address into an integer (so there's no warning; because you are insisting to the compiler that you know what you are doing) but does not dereference the pointer.

Answer 2

You get an error because you didn't actually dereference p, you just casted it (forcefully changed its type) from int*. To dereference, you need to use the * operator, like so:

int i = *p;

Answer 3

You aren't dereferencing the pointer, so you are assigning the address of the pointer to an int.

int i = *p;

Is what you most likely wanted.