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From: hollen@eta.megatek.uucp (Dion Hollenbeck)
Newsgroups: alt.msdos.programmer,comp.sys.ibm.pc
Subject: Re: Turbo C far pointers
Message-ID: <792@megatek.UUCP>
Date: 1 Nov 89 00:54:21 GMT
References: <990@becker.UUCP>
Sender: news@megatek.UUCP
Lines: 45

From article <990@becker.UUCP>, by bdb@becker.UUCP (Bruce Becker):
> In article <565@titan.tsd.arlut.utexas.edu> elrond@titan.tsd.arlut.utexas.edu (Brad Hlista) writes:
> |I am having a problem of getting an allocated block of far memory
> |to return float values.  Here is how variables are declared and dereferenced:
> |
> |far *ptr;
> |
> |main()
> |{
> |int i;
> |
> |	ptr=(float *) farmalloc(100000);
> |
> |	f(i=0;i<1000;i++)
> |		*(ptr+i)=3.1415927;
> 
> Should be:	*((float *)(ptr+i)) = 3.1415927;
> 
> 	Otherwise the "far" declaration of "ptr"
> 	will force a type conversion to the default
> 	of type "int".

If the pointer is declared to be of type "float far *" then the
statement  "ptr + i" says "ptr + sizeof(type of ptr)*i" even
though i is an int.  If this does not occur, then the compiler
is broken.  The program could be written thusly:

main()
{
	float far *ptr;

	ptr = (float *) farmalloc(100000);

	f(i=0;i<1000;i++)
		*(ptr+i)=3.1415927;

}

The key is to declare the pointer to be a far ptr to a type of float
and incrementation and pointer arithmetic will work.

	Dion Hollenbeck             (619) 455-5590 x2814
	Megatek Corporation, 9645 Scranton Road, San Diego, CA  92121

        uunet!megatek!hollen       or  hollen@megatek.uucp