🚀 go-pugleaf

sonix@schunix.uucp (Duane Morin) writes:
> This is driving us NUTS!! Will someone please take a look at the following 
> code and see if they can figure out why it does what it does?  Thank you.
> 
> The following code segment shows a snippet of a program that needs
> to calculate
> the log base 2 of 128 (which happens to be 7).  The problem is that the 
> code works fine on a 386DX/40, with 387 installed, and does NOT work on 
> a 486DX/33!  Help??
> 

I included a shortened version of your program.  I did not completely
solved it but I do have some clues as to what's causing the problem.
After compiling with the assembly output option, I found that at
least two methods can be used to make it return the answer of 7.

One method is to pop the answer off of the co-processor stack and
pop it back on using the fstp and fld instructions prior to the
call to N_FTOL (float to long, near call version). This method
is equivalent to the two step processes of as shown in the code.

The other method is to substitute N_FTOL with the use of fistp
(store and pop stack as integer) and followed by the use of the
lower part of the of the quad word as the result.

I would guess at this point that the function N_FTOL inadvertantly
modifies either the rounding or the precision bits of the
co-processor control word. I will leave it to you and others to
figure out why this happens on the 486 chip and not
on the 386/387 chips, mainly because I need to get back to work.

----------
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <float.h>

// The following constants are normally defined within the program:

#define LOG2 (log(2))
#define LOGN(y)        (log ((double)y) / LOG2 )
main()
{
 	double l;
	int i;

	l = LOGN(128);
	i = (int)l;
	printf("two step = %d\n", i); // this comes up with 7

	i = (int)(LOGN(128));
	printf("one step = %d\n", i); // this comes up with 6?

	return(0);
}

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