[RLUG] C question

[sstrahm4 at charter.net]

Folks-

Because Austin's response and my response contradict, I thought I would tell that Austin's is correct.

The C 2nd Ed (1988) specifies the "new" way of doing declaration lists, and looking at the C History at http://cm.bell-labs.com/cm/cs/who/dmr/chist.html, it shows the old way of making function calls.

I first coded in C in 1984, but did not do any serious coding in C until 1992, after the 2nd Ed came out.

I hate spreading ignorance....

By the way, the 2nd Ed grammar tells that this format remains acceptable.  It would be interesting to know if it is acceptable today.


-Scott

---- Austin Stanhope <stanhope at cse.unr.edu> wrote: 
> 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
> 
> Abuse it and you'll loose it.  I WILL Procmail you.
> Think before you click that forward button. 
> 
> 
> 
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