Re: The FPU emulation problem (2nd time)

   As two of you advised me to use %f instead of %d, i changed the program and
 executed it again. This time I got a different result.  I don't think the FPE
 works at all.  Do you think I can improve this situation by any means?  I app
reciate your advice.

(Changed part)

	fa = fmult(3.0, 5.0);
	printf("float   %f\n", fa);
	da = dmult(3.0, 5.0);
	printf("double  %f\n", da);
	
(New result)
int        15
float     -0.
double   -0.
long      15
==================(previous mail 3/17/99 start)========================
   I succeeded to boot NetBSD on my Mac.  Thank you very much.  Since I have g
ot comp.tgz in addition to base, kern and etc.tgz, I tried to make a program. 
 Although I did my best to execute the program of which the source is shown be
low, it always makes wrong answers.  This makes four answers of the same multi
plication in different types, and two of them are wrong: float and double ones
 become 1076756480, not 15.000000.
   After running another program, I reached a supposition that FPU emulator do
es not work properly (I am using LC475).  It seems to be ready whenever I boot
 my machine, for I can find an message that says "fpu0 at mainbus0 (emulator)"
 before the login prompt appears.  In reality it does not seem to work at all 
when real number calculation is needed.
   I want to know how to have the emulator work correctly.  Do I need anything
 more to improve my system ?  (I do not think so, in fact.)  Please respond to
 me.


(This is the program I ran.)

#include <stdio.h>

main()
{
	int             ia, imult(int, int);
	float           fa, fmult(float, float);
	double          da, dmult(double, double);
	long            la, lmult(long, long);

	ia = imult(3, 5);
	printf("int     %d\n", ia);
	fa = fmult(3.0, 5.0);
	printf("float   %d\n", fa);
	da = dmult(3.0, 5.0);
	printf("double  %d\n", da);
	la = lmult(3l, 5l);
	printf("long     %ld\n", la);
}

int 
imult(int x, int y)
{
	return (x * y);
}
float 
fmult(float x, float y)
{
	return (x * y);
}
double 
dmult(double x, double y)
{
	return (x * y);
}
long 
lmult(long x, long y)
{
	return (x * y);
}


(This is the result of the above program on my Mac.)

int        15
float     1076756480
double   1076756480
long      15
==================(previous mail 3/17/99 end  )========================

Regards,