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On 07/18/2010 06:56 PM, Firas Kraiem wrote:
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> On 18/07/2010 12:27, Aaron Lewis wrote:
>> Hi,
>> 	how can i determine the maximum size of `double' without printing out
>> DOUBLE_MAX ?
>>
>> 	And what about the precision of a `double' , anyone give a try ?
>>
>> 	Google ain't giving much information , i just can't figure out , it's
>> very funny coding , isn't it ?
>>
> 
> Kinda off-topic for here but whatever...

Yeah , i'll put a OT here.

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> 
> Assuming 64-bit double, it has a mantissa of 52 bits and an exponent of
> 11 bits. If the mantissa is all-ones, that will give a significant of
> 2-2^(-52).  The exponent cannot be all-ones (an all-ones exponent means
> infinity or NaN), so the largest exponent is 2^11-2 = 2046, minus bias
> (1023), it gives a maximum effective exponent of 1023, so that would
> give a value of (2-2^(-52))*2^(1023) = 2^1024 - 2^971, roughly 10^308.

Yep , right , but how you *get* it by a beautiful code  , that could be
really interesting.

I mean , with a small code , c or c++ , printing the size out.

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> 
> What do you mean by "precision"?

Kinda of significance digit. For example:
 A number `12.340' , and if say it has 2 digits' precision , then we
consider the `0' is not accurate , while `.34' is accurate.

So for a number stored in a double type , how accurate can it be ? (or
maybe how many bits in the fixed-point part is accurate)

Doesn't matter if it's unsigned or signed , some ideas are cool enough.

Still , use a small part of code , c or c++ , i'm just curious how to
make it happen.

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> 
> Firas


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