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From: karl@haddock.UUCP (Karl Heuer)
Newsgroups: comp.lang.c
Subject: Re: Copying a constant number of bytes
Message-ID: <563@haddock.UUCP>
Date: Mon, 15-Jun-87 17:15:15 EDT
Article-I.D.: haddock.563
Posted: Mon Jun 15 17:15:15 1987
Date-Received: Sun, 21-Jun-87 02:50:25 EDT
References: <900@bloom-beacon.MIT.EDU>
Reply-To: karl@haddock.ISC.COM.UUCP (Karl Heuer)
Distribution: world
Organization: Interactive Systems, Boston
Lines: 32

In article <900@bloom-beacon.MIT.EDU> newman@athena.mit.edu (Ron Newman) writes:
>[To copy a constant number of bytes one can use bcopy/memcpy, or]
>      struct nbytes {char s[NBYTES];};
>      *(struct nbytes *)b = *(struct nbytes *)a;
>... But can I count on a compiler generating efficient code for method (2)?

Well, you can't count on a compiler generating efficient code for ANYTHING,
but the compiler will almost certainly use something at least as efficient
(timewise) as the bcopy/memcpy subroutine.  (Usually it's better, since the
compiler knows the exact size, whereas the subroutine has to handle the
general case.)

Unfortunately, being efficient isn't as important as being right.  The major
problem is that the pointers a and b, since they are not really of this struct
type, may not have the correct alignment.  In particular, I seem to recall the
SVR2 compiler for the 3B2 would force a stricter-than-necessary alignment (and
hence size) on all structs.  Thus, you might even get more bytes than you
asked for.

>Is the answer any different if I know that "a" and "b" are 32-bit aligned, or
>that NBYTES is a multiple of 4?

The answer to your original question is "No", unless the compiler knows it
too.  (It does know the value of NBYTES, but probably doesn't know the
alignment of the pointers.)  The answer I gave is still correct (it's not
strictly portable), but if the size and alignment are sufficiently large you
happen to get the right answer on the machines I'm familiar with.

My recommendation is to use memcpy.  If you choose not to, I suggest that the
resulting code should be well-commented.

Karl W. Z. Heuer (ima!haddock!karl or karl@haddock.isc.com), The Walking Lint