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.Java@[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.Java@[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
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