[Bf-committers] Verse 2005 Errors
branan at gmail.com
Thu Sep 14 22:41:10 CEST 2006
alright... 32-bit float has a 1-bit sign, 8-bit exponent and 23-bit
OK. the 8-bit exponent gives us a maximum power of 127, since there's
a bias to allow for negative exponents.
The maximum floating point number is normalized, so it's
1.xxxxxxxxxxxxxxxxxxxxxxx, which is a 24-bit number with the first bit
always set. Since we're going for the maximum number, that value is
all ones: 1.11111111111111111111111
That value is 16777215.
Floating point numbers = significand * 2^exponent. In this case, that
should be 1677215 * 2^127. which is showing up as inf because it's
been rounded up. Try changing that last 2 to a 1, and see what
On 9/14/06, Emil Brink <emil at obsession.se> wrote:
> Branan Riley wrote:
> > 16,777,215 * 2^127.
> Hm ... I don't think that is the correct interpretation of the bit pattern
> as a float.
> > Running that through Kcalc, I get 2.854495215270736302e+45. I'd do it out
> > by hand to get an exact number, but that's a 152-bit number, so I'm not
> > going to even think about it. If someone wants to run that through, and
> > see what comes up for the binary representation, that would be nifty.
> Okay ... I think you lost me, though. I'm expecting the float literal
> to have a decimal exponent of +38, that part has never been in doubt.
> If the proper value has an exponent of +45, I'd be ... very surprised.
> A quick test with the above value:
> ~> gcc -x c -
> #include <stdio.h>
> int main(void)
> float a = 2.854495215270736302e+45;
> printf("a=%g\n", a);
> return 0;
> ~> ./a.out
> Seems to show that you've lost not only me, but the C compiler as well.
> Can you run that by me again?
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