CODEWITHSUNDEEP
c
jaz_n
jaz_n

Reputation: 41

Why would a function use the "address of" operator on an argument passed to it as a pointer?

I am new to the C programming language and I am trying to understand the intent of a function which takes a pointer argument but uses the address of operator on it as though it was a variable name in the function's body? The parameters being passed are of type struct. In other words, why did the author choose to use &lib -> fatfs; instead of lib->fatfs? Is the "address of" operator used to ensure that a null pointer is not being passed?

void SDCardLib_init(SDCardLib * lib, SDCardLib_Interface * interface)
{
    lib->interface = interface;

    f_mount(0, &lib->fatfs);

    disk_initCallBack(interface);
}

Upvotes: 1

Views: 72

Answers (2)

Petr Skocik
Petr Skocik

Reputation: 60068

&lib->fatfs is the same as &(lib->fatfs) or &((*lib).fatfs). The & doesn't take the address of the pointer named lib.

&lib->fatfs is essentially whatever lib points at plus whatever offset the fatfs member is at.

The following piece of code asserts that:

//imagine a fake SDCardLib
typedef long sometype;
typedef struct {
    int something0;
    sometype fatfs;
    int something1;
} SDCardLib;

#undef NDEBUG //make sure asserts are kept
#include <assert.h>
#include <stddef.h> //offsetof
#include <stdlib.h> //malloc
int main()
{
    SDCardLib *lib = malloc(sizeof *lib);
    if(!lib) return 1;

    assert( &lib->fatfs  ==
                (sometype*) ((char*)lib + offsetof(SDCardLib,fatfs)) );

    assert( &lib->fatfs == &(lib->fatfs) );
    assert( &lib->fatfs == &((*lib).fatfs) );
}

Upvotes: 4

dbush
dbush

Reputation: 223917

The address-of operator is not being applied to the pointer lib but to the fatfs member of the pointed to object. The -> operator has higher precedence then the unary & operator.

That means that &lib->fatfs is the same as &(lib->fatfs). This should make it more clear what & is taking the address of.

Upvotes: 5

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