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June 18, 2006 How many bits in the mantissa if I declare all floats as long doubles ? | ||||
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The source code for an entrophy testing method I am trying to incorporate in my program has the comment: /* Bytes used as Monte Carlo co-ordinates. This should be no more bits than the mantissa of your "double" floating point type. */ #define MONTEN 6 I cannot determine from either the .asm or the .h or the help files how many bits are in the mantissa. |
June 18, 2006 Re: How many bits in the mantissa if I declare all floats as long doubles ? | ||||
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Posted in reply to Nicholas Jordan | Nicholas Jordan wrote:
> The source code for an entrophy testing method I am trying to incorporate in my
> program has the comment:
>
> /* Bytes used as Monte Carlo co-ordinates.
> This should be no more bits than the mantissa
> of your "double" floating point type. */
>
> #define MONTEN 6
>
> I cannot determine from either the .asm or the .h or the help files how many
> bits are in the mantissa.
long doubles have a 64 bit mantissa, a 15 bit exponent, and a 1 bit sign.
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June 19, 2006 Re: How many bits in the mantissa if I declare all floats as long doubles ? | ||||
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Posted in reply to Walter Bright | Walter Bright wrote:
> Nicholas Jordan wrote:
>> The source code for an entrophy testing method I am trying to
>> incorporate in my
>> program has the comment:
>>
>> /* Bytes used as Monte Carlo co-ordinates.
>> This should be no more bits than the mantissa
>> of your "double" floating point type. */
>>
>> #define MONTEN 6
>>
>> I cannot determine from either the .asm or the .h or the help files
>> how many
>> bits are in the mantissa.
>
>
> long doubles have a 64 bit mantissa, a 15 bit exponent, and a 1 bit sign.
I'd really like to see a SEF float format with two or more sign bits. :-)
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