Simple Question on Pointers

Question

Can someone remind me why this works?

A function requiring int* could take as input an (obviously)

int *integer;

but it could also accept

&var->integer

with var being var_t*, where var_t:

typedef struct {
    int integer;
} var_t;

why is the 2nd accepted?

edit: oopsy, question is same but var is actually a var_t* (and not a var_t) to be more precise.

Answer 1

The second version is accepted because of the ampersand at the beginning. This means that the address of the field is passed, not the actual value.

Answer 2

2nd version of code is only good version, because you must use & operator to get address of integer in var_t structure.

Answer 3

Let's break it down.

var is a var_t*.

var->integer is an int.

&var->integer is an int*.

Answer 4

If the type of var is var_t, the code you've shown is actually illegal. If its type is var_t*, it's accepted because the type of var->integer is int and the type of &some_int is int*.

Answer 5

Do the math your self:

int a => &a is of type int*
int *a => &a is of type int**
int **a => &a is of type int***

etc.

int *a => *a is of type int
int **a => a is of type int**
int **a => *a is of type int*
int **a => **a is of type int