--- Dan Erickson <coldoneknight@rogers.com> wrote:
>
<snip>
> Basicly when your writing... how shall I say....?? *normal code*,
> you use
> your functions in such a manner as follows.
>
> int foo;
> char *bar = "5";
>
> foo = atoi(bar);
>
> now here we give our funtion an argument
>
> while when you have kernel code, instead of writing the actual
> args, you
> basicly make up your own function like...
>
> static ssize_t wdt_write(struct file *file, const char *buf, size_t
> count,
> loff_t *ppos)
> {
> /* Can't seek (pwrite) on this device */
> if (ppos != &file->f_pos)
> return -ESPIPE;
>
> if(count)
> {
> foo();
> return 1;
> }
> return 0;
> }
>
>
> What has me confussed is.... what gives this function the arguments
> so it
> can function properly. Ie) as bar is the arg to atoi() earlier.
>
Let me make sure that I understand you correctly:
In the example above there is a function by the name of "wdt_write".
This function returns a value of type "ssize_t", and is passed four
arguments:
struct file *file,
const char *buf,
size_t count,
loff_t *ppos.
Your question, if I am understanding you, is: when function
wdt_write is executed, where does it get its passed arguments?
If this is your question, here is the answer:
It gets its arguments from the same place that atoi() in your first
example does; namely, from the code that *calls* it.
Hope this was in any way useful.
--Christine
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