[RLUG] C question

[Austin Stanhope]

It's an older C way of giving types to the function arguments.  It has 
since been changed so you can do it the old way, or the way most of us 
are used to doing it (the term most is open to interpretation here).  It 
is equivalent to saying:

int timeval_subtract (timeval *result, timeval *x, timeval *y)
{
    // do stuff
}

-Austin



Grant Kelly wrote:
> I came across this code which does something I've never seen before. 
> Could someone explain what the variable declaration between the 
> function header and the function body do? Or at least point me in a 
> direction to find out for myself. Again, I know what the code does, 
> but I don't know what the line marked with <==== does.
>
> Thanks,
> Grant
>
> int timeval_subtract (result, x, y)
>      struct timeval *result, *x, *y;    <==== What does this do???
> {
>   /* Perform the carry for the later subtraction by updating y. */
>   if (x->tv_usec < y->tv_usec) {
>     int nsec = (y->tv_usec - x->tv_usec) / 1000000 + 1;
>     y->tv_usec -= 1000000 * nsec;
>     y->tv_sec += nsec;
>   }
>   if (x->tv_usec - y->tv_usec > 1000000) {
>     int nsec = (y->tv_usec - x->tv_usec) / 1000000;
>     y->tv_usec += 1000000 * nsec;
>     y->tv_sec -= nsec;
>   }
>
>   /* Compute the time remaining to wait.
>      tv_usec is certainly positive. */
>   result->tv_sec = x->tv_sec - y->tv_sec;
>   result->tv_usec = x->tv_usec - y->tv_usec;
>
>   /* Return 1 if result is negative. */
>   return x->tv_sec < y->tv_sec;
> }
>
> ------------------------------------------------------------------------
>
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-- 
Austin Stanhope                      www.cse.unr.edu/~stanhope
Robotics Research                    www.cse.unr.edu/~society
Computer Science and Engineering     www.cse.unr.edu
University of Nevada, Reno           www.unr.edu

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