malloc() and sizeof

[Piet van Oostrum]

In article <624 at gonzo.UUCP>, daveb at gonzo (Dave Brower) writes:
 `However, the real case is quite often:
 `
 `	struct foo *tmp;
 `
 `	/* tons-o-code deleted */
 `
 `	tmp  = (struct foo *)malloc( /*your choice here*/ );
 `
 `It is all too easy to forget the type of "tmp" at this point.  If you do
 `your malloc argument as sizeof(*tmp), you cannot go wrong.  Further, if
 `you change the type of the definition of tmp above, you will not need to
 `change the argument to sizeof.
 `
 `
 `Of course, this argument bogs down when you realize you already needed
 `to know the type to get the cast on the malloc return correct.  But
 `given the choice, I'd still rather only have to get the type right once
 `(in the cast) rather than in the cast _and_ the sizeof.

The following example has a cast that is different from the type of the
variable:

	typedef ..... FOO;
	FOO buf [BUFSIZ];
	FOO *bufcopy;
	....
	bufcopy = (FOO *) malloc (sizeof (buf));
	bcopy (bufcopy, buf, sizeof (buf));

Of course you can use sizeof (FOO[BUFSIZ]), but why would you. It is much
cleaner (IMHO) to use sizeof buf. This esample won't happen very often in
practice, I think, but it is better to use a single programming style.
-- 
Piet van Oostrum, Dept of Computer Science, University of Utrecht
Padualaan 14, P.O. Box 80.089, 3508 TB Utrecht, The Netherlands
Telephone: +31-30-531806. piet at cs.ruu.nl (mcvax!hp4nl!ruuinf!piet)