11  2349 
spasmous wrote:
Just wondering.
About what?
Ah, I see, please put your question in the body of your message.
Can a double be 32 bits?
No, 32 bits is too small.
--
Ian Collins.
On Apr 18, 3:19*pm, Ian Collins <ian-n...@hotmail.co mwrote:
spasmous wrote:
Just wondering.
About what?
Ah, I see, please put your question in the body of your message.
Can a double be 32 bits?
No, 32 bits is too small.
I can't locate anywhere official-looking that it says this. Does the C
standard mandate some minimum number of bits?
"spasmous" <sp******@gmail .comwrote in message
news:15******** *************** ***********@p25 g2000pri.google groups.com...
On Apr 18, 3:19 pm, Ian Collins <ian-n...@hotmail.co mwrote:
>spasmous wrote:
Just wondering.
About what?
>Ah, I see, please put your question in the body of your message.
>Can a double be 32 bits?
>No, 32 bits is too small.
>I can't locate anywhere official-looking that it says this. Does the C
standard mandate some minimum number of bits?
Aparently the minimum is 10 decimal digits. With the exponent added, 32 bits
is not enough.
In practice a double will likely be 64 bits. But with C nothing is ever
certain.
--
Bart
In article <15************ *************** *******@p25g200 0pri.googlegrou ps.com>,
spasmous <sp******@gmail .comwrote:
>On Apr 18, 3:19=A0pm, Ian Collins <ian-n...@hotmail.co mwrote:
>spasmous wrote:
Just wondering.
>About what?
>Ah, I see, please put your question in the body of your message.
>Can a double be 32 bits?
>No, 32 bits is too small.
>I can't locate anywhere official-looking that it says this. Does the C
standard mandate some minimum number of bits?
No, but it mandates minimum value ranges and minimum precisions.
About a week ago, there was a calculation posted showing that
approximately 40.4 value bits would be required to meet the specification;
with 1 "hidden bit", it is perhaps doable in 40 bits. But not 32.
--
"Man's life is but a jest,
A dream, a shadow, bubble, air, a vapor at the best."
-- George Walter Thornbury
"Bartc" <bc@freeuk.comw rote in message
news:Jt******** **********@text .news.virginmed ia.com...
>
"spasmous" <sp******@gmail .comwrote in message
news:15******** *************** ***********@p25 g2000pri.google groups.com...
On Apr 18, 3:19 pm, Ian Collins <ian-n...@hotmail.co mwrote:
>>spasmous wrote:
Just wondering.
About what?
>>Ah, I see, please put your question in the body of your message.
>>Can a double be 32 bits?
>>No, 32 bits is too small.
>>I can't locate anywhere official-looking that it says this. Does the C
standard mandate some minimum number of bits?
Aparently the minimum is 10 decimal digits. With the exponent added, 32
bits is not enough.
Even without the exponent, it takes ~33.22 bits to maintain correctly 10
decimal digits, and we also need to be able to store an exponent of +/-37,
requiring ~5.2 bits so 39 bits +2 bits for signs (one implicit) gives us 40
bits as a mathematical bare minimum that we could expect to be used for data
type double in the C language.
In practice a double will likely be 64 bits. But with C nothing is ever
certain.
A double will have at least 10 decimal digits of precision and an exponent
of at least +/- 37. So at least that much is certain.
;-)
** Posted from http://www.teranews.com **
Ian Collins <ia******@hotma il.comwrites:
spasmous wrote:
>Just wondering.
About what?
The "Subject"?
Tricky eh?
>
Ah, I see, please put your question in the body of your message.
Can a double be 32 bits?
No, 32 bits is too small.
On 19 Apr, 00:55, "Dann Corbit" <dcor...@connx. comwrote:
"Bartc" <b...@freeuk.co mwrote in message
....
>No, 32 bits is too small.
>I can't locate anywhere official-looking that it says this. Does the C
standard mandate some minimum number of bits?
Aparently the minimum is 10 decimal digits. With the exponent added, 32
bits is not enough.
Even without the exponent, it takes ~33.22 bits to maintain correctly 10
decimal digits, and we also need to be able to store an exponent of +/-37,
requiring ~5.2 bits so 39 bits +2 bits for signs (one implicit) gives us 40
bits as a mathematical bare minimum that we could expect to be used for data
type double in the C language.
5.2 bits??? I think the exponent is stored in binary so would need to
be larger. Assuming the exponent can be in the range
from: 10 ** -37
to : 10 ** 37
since
log2(10 ** -39) = -129.55519570060 713
log2(10 ** -38) = -126.23326760571 979
log2(10 ** -37) = -122.91133951083 242
log2(10 ** 37) = 122.91133951083 242
log2(10 ** 38) = 126.23326760571 976
log2(10 ** 39) = 129.55519570060 713
the exponent seems to need a value in the range -123 up to +123 (247
values) and therefore would need 8 bits. Interestingly it looks like
the range for 10 ** +/- 38 could just be accommodated in eight bits as
it needs -127 up to 127 (255 values) leaving one value to indicate
others such as NaNs.
On Apr 21, 8:29*am, James Harris <james.harri... @googlemail.com >
wrote:
On 19 Apr, 00:55, "Dann Corbit" <dcor...@connx. comwrote:
"Bartc" <b...@freeuk.co mwrote in message
...
>>No, 32 bits is too small.
>>I can't locate anywhere official-looking that it says this. Does the C
>>standard mandate some minimum number of bits?
Aparently the minimum is 10 decimal digits. With the exponent added, 32
bits is not enough.
Even without the exponent, it takes ~33.22 bits to maintain correctly 10
decimal digits, and we also need to be able to store an exponent of +/-37,
requiring ~5.2 bits so 39 bits +2 bits for signs (one implicit) gives us40
bits as a mathematical bare minimum that we could expect to be used for data
type double in the C language.
5.2 bits??? I think the exponent is stored in binary so would need to
be larger. Assuming the exponent can be in the range
from: 10 ** -37
to *: 10 ** 37
since
log2(10 ** -39) = -129.55519570060 713
log2(10 ** -38) = -126.23326760571 979
log2(10 ** -37) = -122.91133951083 242
log2(10 ** 37) = 122.91133951083 242
log2(10 ** 38) = 126.23326760571 976
log2(10 ** 39) = 129.55519570060 713
the exponent seems to need a value in the range -123 up to +123 (247
values) and therefore would need 8 bits.
If you read carefully, you will see that I reserved an extra bit for
the sign.
pow(2,
5.2094533656289 497818578041776 131754997614249 696721880487516 636221416853906 4)
is ~37
Hence, 5.2 bits.
Interestingly it looks like
the range for 10 ** +/- 38 could just be accommodated in eight bits as
it needs -127 up to 127 (255 values) leaving one value to indicate
others such as NaNs.
The C standard only asks for +/-37 and so trying to obtain 38 is not
salient.
It would be possible to physically store floating numbers using
artithmetic compression, so fractional bits are possible.
On 21 Apr, 21:04, user923005 <dcor...@connx. comwrote:
....
>No, 32 bits is too small.
>I can't locate anywhere official-looking that it says this. Does the C
>standard mandate some minimum number of bits?
Aparently the minimum is 10 decimal digits. With the exponent added, 32
bits is not enough.
Even without the exponent, it takes ~33.22 bits to maintain correctly 10
decimal digits, and we also need to be able to store an exponent of +/-37,
requiring ~5.2 bits so 39 bits +2 bits for signs (one implicit) gives us 40
bits as a mathematical bare minimum that we could expect to be used for data
type double in the C language.
5.2 bits??? I think the exponent is stored in binary so would need to
be larger. Assuming the exponent can be in the range
from: 10 ** -37
to : 10 ** 37
since
log2(10 ** -39) = -129.55519570060 713
log2(10 ** -38) = -126.23326760571 979
log2(10 ** -37) = -122.91133951083 242
log2(10 ** 37) = 122.91133951083 242
log2(10 ** 38) = 126.23326760571 976
log2(10 ** 39) = 129.55519570060 713
the exponent seems to need a value in the range -123 up to +123 (247
values) and therefore would need 8 bits.
If you read carefully, you will see that I reserved an extra bit for
the sign.
Sure (though an exponent is normally stored biased and non-negative so
doesn't have a sign bit per se) but neither 5, 5.2, nor (with an extra
'sign' bit) 6 or 6.2 bits seem to be enough. I was saying that (if the
mantissa is a binary number as normal) the exponent will need to range
from -123 to +123, not -37 up to +37. (2 ** 123 is a little over 10 **
37 to cover the exponent of +/- 37 that you mention is needed for C) I
think you therefore need 8 bits for the exponent for the C requirement
of 10 ** +/- 37.
You could possibly make the mantissa a decimal number and then use
your 5.2 or 6.2 bits figure but a decimal number would not be stored
as efficiently as a binary one and would therefore store less in a
given space.
....
Interestingly it looks like
the range for 10 ** +/- 38 could just be accommodated in eight bits as
it needs -127 up to 127 (255 values) leaving one value to indicate
others such as NaNs.
The C standard only asks for +/-37 and so trying to obtain 38 is not
salient.
My comment was the other way round, i.e. given that an 8-bit exponent
is needed (as it is to hold the 247 values from -123 up to 123) it can
hold higher numbers using some of the 9 remaining values (256 - 247 =
9). At least one of the values needs to be reserved for NaNs etc but
that still seems to allow 10 ** +/- 38. It would depend on the
representation.
It would be possible to physically store floating numbers using
artithmetic compression, so fractional bits are possible.
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