On Mar 22, 4:34 pm, Lew <l...@[EMAIL PROTECTED]
> wrote:
> Patricia Shanahan wrote:
> > Perhaps you could identify a bit pattern in that range that does not
> > represent a floating point number, or a bit pattern outside the range
I
> > gave that corresponds to a floating point number in 0f through 1f.
> Alex.From.Ohio.J...@[EMAIL PROTECTED]
wrote:
> >> So, (don't forget normalization, which actually prevents such things,
> >> but the DO exist in memory) we need to know what are
> >> exponent (2*e) and fraction (1.+ .f) in our implementation.
> >> In Java float is 32 bites.
> >> So, e has 8 bits and f has 23.
>
> So there are 31 bits, plus the sign bit.
>
> Alex.From.Ohio.J...@[EMAIL PROTECTED]
wrote:
> >> Anything which is less then half of 2 is 1. So, All e<0 (127) are
> >> between 0 and 1.
> >> That's 2**23 * 127
>
> 2 ** 23 * (2 ** 7 - 1 )
> = 2 ** 30 - 2 ** 23
>
> Your calculations already are on the order of 2 ** 30.
>
> Patricia Shanahan wrote:
> > Normalized floats in IEEE754 do not store the leading 1 bit. However,
> > the positive denomalized floats, bit patterns 0x00000001 through
> > 0x007fffff, are also in the specified range.
>
> --
> Lew
OK, OK.
Little tiny hint doesn't work.
Let's use brutal force:
float f;
long i=0, c=0;
for(; (i&0x100000000L)==0; i++){
f=Float.intBitsToFloat((int)i);
if (f<=2f&&f>=0f){
c++;
}
}
System.out.println(c+" "+i);
System.out.println(Math.pow(2, 30));
and output is:
1073741826 4294967296
1.073741824E9
As you can see float range from 0 to 1 inclusively is 2**30 + 2
(which are -0 and 1).
Sincerely, Alex.
http://www.myjavaserver.com/~alexfromohio/
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