From:	George Foot <mert0407 AT sable DOT ox DOT ac DOT uk>
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: Sizeof and pointers
Date:	5 Feb 1998 08:33:19 GMT
Organization:	Oxford University, England
Lines:	59
Message-ID:	<6bbtgf$3v0$6@news.ox.ac.uk>
References:	<34D8C88E DOT 2849A0A AT cs DOT curtin DOT edu DOT au>
NNTP-Posting-Host:	sable.ox.ac.uk
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To:	djgpp AT delorie DOT com
DJ-Gateway:	from newsgroup comp.os.msdos.djgpp

On Thu, 05 Feb 1998 03:59:10 +0800 in comp.os.msdos.djgpp David
Shirley <shirleyd AT cs DOT curtin DOT edu DOT au> wrote:

: void func(unsigned char *tmp)
: {
: // BLAH BLAH
: }

: How do i find how many bytes tmp is taking up in memory:

: the sizeof funtion returns 4 bytes, the reason being (i think) that it
: is finding out how many bytes the pointer takes up in memory. I dont
: want to use strlen(tmp) because the string may have a '\0' character.

`tmp' *is* a pointer, and sizeof (tmp) is telling you exactly how big
it is.  What you want is a way of finding the size of the thing `tmp'
points at.  You can use `sizeof (*tmp)' to find this -- you'll get 1,
since `tmp' points at a char.  Not very useful, eh?

The simple answer is that there is no way your function can know,
given only the value of `tmp'.  You'll have to do some better
bookkeeping, I'm afraid.  You imply that `tmp' is pointing at a
string; if you guarrantee that the string is properly formed you won't
have any problem here.  If null-terminating it C-style is problematic,
perhaps using a Pascal-style string (where tmp[0] is the length of the
string, and tmp[1] is the first character) would work better.

If it is likely not to be a proper text string (perhaps it's just a
block of binary data) then you'll have to pass a second parameter to
the function, giving the size of the object.  Surely the caller knows
how big the block is?  Some part of your program must have allocated
it :).

As a separate note, if what is being pointed at is an array then you
can find out how large it is, using `sizeof' as normal:

---- start -----
#include <stdio.h>

int array[10];

int main (void) {
	int a = sizeof (array);
	int b = sizeof (*array);
	printf ("sizeof (array)  = %d\n", a);
	printf ("sizeof (*array) = %d\n", b);
	printf ("num of elements = %d\n", a/b);
	return 0;
}
----- end -----

This is a pure C question, and shouldn't really have been asked on the
djgpp newsgroup -- comp.lang.c would probably have been a more
suitable place.

-- 
george DOT foot AT merton DOT oxford DOT ac DOT uk

Remember what happened to the dinosaurs.

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