🚀 go-pugleaf

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". The way you have it the
	contents of of *(ptr+i)" contains an int
	value which is coerced back to float
	in the printf statement.

|	for(i=0;i<10;i++)
|		printf(" %f", (float) *(ptr+i) ) ;  /* is yielding 3.00000 */

Should be	printf(" %f", *((float *)(ptr+i)) ) ;  /* is yielding 3.1415927 */

|}
|
|Can someone please help me understand what is going on?
|Thanks.

	You need to ensure that either by default
	or by explicit declaration that
	sizeof(far) == sizeof(float); is there a
	"far long" type? Better yet, a "far float"?

	I didn't actually try this so if your compiler
	barfs on "*((float *)ptr)" where "ptr" is a
	"far *" declaration, well... you could try
	"far float * ptr;", etc...

Cheers,
-- 
  .::.	 Bruce Becker	Toronto, Ont.
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