On 10/4/06, Hendrik Post <hendrik@xxxxxxxxxxxx> wrote:
Hi Ricard, Thank you for your answer. My problem was my incorrect (?) understanding of "s->foo". I thought it being equivalent to "(*s).foo" rather than a direct offset calculation.
They are equivalent. The rule is simple, no value is read unless a read operation is performed. Also, no value is written unless a write operation is performed. When you write a = b; read operation is required on 'b', and a write on 'a'. But when you say a; // no operation is specified here or ; // an empty statement neither read nor a write is asked to perform by us. The compiler will thus silently ignore such a statement since we have not asked for any particular operation over 'a'. A read or write operation has two steps: Step 1: Get the address of the variable upon which to operate on Step 2: Read/Write to that address Now, when you write a = &b; both the above steps are performed over 'a' whereas only step 1 is performed over 'b'. It's because we've not asked the compiler to do that. Therefore when you say a = &(s->foo); the possible steps are: Step 1: Get the address of s Step 2: Read the value(address) contained in s Step 3: Get the address of member foo inside s Step 4: Read the value of foo Step 4 is not performed because we've already obtained our required result in Step 3 or in other words - we've not asked the compiler to read the value of foo; only the address of it. I think I've explained too much, Best of Luck, Jinesh. -- Kernelnewbies: Help each other learn about the Linux kernel. Archive: http://mail.nl.linux.org/kernelnewbies/ FAQ: http://kernelnewbies.org/faq/
