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From: brnstnd@kramden.acf.nyu.edu (Dan Bernstein)
Newsgroups: comp.lang.c
Subject: Re: A quick question...
Message-ID: <20125:Mar1209:03:3491@kramden.acf.nyu.edu>
Date: 12 Mar 91 09:03:34 GMT
References: <1991Mar12.030759.26698@nntp-server.caltech.edu>
Organization: IR
Lines: 37

In article <1991Mar12.030759.26698@nntp-server.caltech.edu> eychaner@suncub.bbso.caltech.edu writes:
> 	unsigned char *pointer1;
> 	short short_value;

The value that ``pointer1'' refers to has not yet been defined. It's
probably garbage.

> 	...
> 	*((short *) pointer1) = short_value;

If ... doesn't include any initialization of pointer1, this has
undefined behavior.

If ... initializes pointer1 to point to some memory location other than
that occupied by a short, your code sequence has undefined behavior.

If ... initializes pointer1 to point to the memory occupied by a short,
then this will work. You can cast freely between char *, void *, and any
(single) other pointer-to-object type. This would work:

  char *pointer1;
  short foo;
  short bar;
  ...
  pointer1 = (char *) &foo;
  *((short *) pointer1) = bar;

This has the same effect as foo = bar. I believe unsigned char * works
the same as char * for this purpose.

> And does it do what I think it does, that is, assign short_value to the
> storage pointed to by pointer1?

Only if you've set pointer1 to point to some valid short storage. (Of
course, you need to initialize short_value as well.)

---Dan