Re: A tough C construct in Linux Kernel's implementation

["Rajat Jain" <rajat.noida.india@xxxxxxxxx>]

>
> #define offsetof(TYPE, MEMBER) ((size_t) &((TYPE *)0)->MEMBER)
>
> is returning the "Offset of a Memeber inside a struct/class Type". I
> didn't logically understand how is that happenning...:-(
>
> I expect the outcome of &((TYPE *)0)->MEMBER to be the address of the
> member 'MEMBER' of type 'TYPE'. Now, how does that address becomes
> offset has totally gone out of my understanding...
>

You are right ... &((TYPE *)0)->MEMBER is the address of member
"MEMBER" of type "TYPE". But consider that the (base) address of the
structure of type "TYPE" is NULL (0) as shown in above expression, and
hence the address of "MEMBER" is nothing but the offset.

An example:

Say you have:

struct student {
  int roll no;
  int age;
}

if you write offsetof(student, age), it converts to:

((size_t) &((student *)0)->age)

= ((size_t) &((student*)NULL)->age)

= ((size_t) 4 )    (Considering int = 4 bytes)

=4 = offset

Regards,

Rajat
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