# Can a double always represent an int exactly?

Discussion in 'C Programming' started by Fred Ma, Oct 22, 2004.

1. ### Fred MaGuest

I'm using the expression "int a = ceil( SomeDouble )".
The man page says that ceil returns the smallest
integer that is not less than SomeDouble, represented
as a double. However, my understanding is that a
double has nonuniform precision throughout its value
range. Will a double always be able to exactly
represent any value of type int? Could someone please
point me to an explanation of how this is ensured,
given that the details of a type realization varies
with the platform?

Thanks.

Fred

P.S. I am not worried about overflowing the int value
range, just about the guaranteed precise representation
of int by double.

Fred Ma, Oct 22, 2004

2. ### Rouben RostamianGuest

In article <>,
Fred Ma <> wrote:
>I'm using the expression "int a = ceil( SomeDouble )".
>The man page says that ceil returns the smallest
>integer that is not less than SomeDouble, represented
>as a double. However, my understanding is that a
>double has nonuniform precision throughout its value
>range. Will a double always be able to exactly
>represent any value of type int? Could someone please
>point me to an explanation of how this is ensured,
>given that the details of a type realization varies
>with the platform?

I don't know whether the C Standard specifies anything to
this effect. But here is an implementation-specific
observation.

On a machine with 64-bit doubles which follow the IEEE
specification, the mantissa part is 53 bits (plus one hidden
bit as well) therefore integers as large as around 2 to the
50th power should be exactly representable. In particular, if
the machine has 32-bit ints, they are all exactly representable
as doubles.

On my machine, which has 32-bit ints and 64-bit doubles,
the following yields the exact answer:

printf("%30.15f\n", 1.0 + pow(2.0, 52.));

However the following stretches it too far and the answer
is inexact:

printf("%30.15f\n", 1.0 + pow(2.0, 53.));

--
rr

Rouben Rostamian, Oct 22, 2004

3. ### Fred MaGuest

Rouben Rostamian wrote:
>
> I don't know whether the C Standard specifies anything to
> this effect. But here is an implementation-specific
> observation.
>
> On a machine with 64-bit doubles which follow the IEEE
> specification, the mantissa part is 53 bits (plus one hidden
> bit as well) therefore integers as large as around 2 to the
> 50th power should be exactly representable. In particular, if
> the machine has 32-bit ints, they are all exactly representable
> as doubles.
>
> On my machine, which has 32-bit ints and 64-bit doubles,
> the following yields the exact answer:
>
> printf("%30.15f\n", 1.0 + pow(2.0, 52.));
>
> However the following stretches it too far and the answer
> is inexact:
>
> printf("%30.15f\n", 1.0 + pow(2.0, 53.));

I realize that if a double actually uses twice as many bits as
ints, the mantissa should be big enough that imprecision should
never arise. I'm just concerned about whether this can be relied
upon. My faith in what seems normal has been shaken after finding
that long has the same number of bits as int in some environments.
What if double has the same number of bits as ints in some
environments? Some of those bits will be taken up by the
exponent, and the mantissa will actually have fewer bits than an
int. Hence, it will be less precise than ints within the value
range of ints.

Fred

Fred Ma, Oct 22, 2004
4. ### Chris TorekGuest

In article <>
Fred Ma <> writes:
>I'm using the expression "int a = ceil( SomeDouble )". The man
>page says that ceil returns the smallest integer that is not less
>than SomeDouble, represented as a double. However, my understanding
>is that a double has nonuniform precision throughout its value range.

This is correct (well, I can imagine a weird implementation that
deliberately makes "double"s have constant precision by often
wasting a lot of space; it seems quite unlikely though).

Note that ceil() returns a double, not an int.

>Will a double always be able to exactly represent any value of
>type int?

This is implementation-dependent. If "double" is not very precise
but INT_MAX is very large, it is possible that not all "int"s can
be represented. This is one reason ceil() returns a double (though
a small one at best -- the main reason is so that ceil(1.6e35) can
still be 1.6e35, for instance).

>Could someone please point me to an explanation of how this is ensured,
>given that the details of a type realization varies with the platform?

I am not sure what you mean by "this", especially with the PS:

>P.S. I am not worried about overflowing the int value
>range, just about the guaranteed precise representation
>of int by double.

.... but let me suppose you are thinking of a case that actually occurs
if we substitute "float" for "double" on most of today's implementations.
Here, we get "interesting" effects near 8388608.0 and 16777216.0.
Values below 16777216.0 step by ones: 8388608.0 is followed
immediately by 8388609.0, for instance, and 16777215.0 is followed
immediately by 16777216.0. On the other hand, below (float)(1<<23)
or above (float)(1<<24), we step by 1/2 or 2 respectively. Using
nextafterf() (if you have it) and variables set to the right values,
you might printf() some results and find:

nextafterf(8388608.0, -inf) = 8388607.5
nextafterf(16777216.0, +inf) = 16777216.2

So all ceil() has to do with values that are at least 8388608.0
(in magnitude) is return those values -- they are already integers.
It is only values *below* this area that can have fractional
parts.

Of course, when we use actual "double"s on today's real (IEEE style)
implementations, the tricky point is not 2-sup-23 but rather
2-sup-52. The same principal applies, though: values that meet or
exceed some magic constant (in either positive or negative direction)
are always integral, because they have multiplied away all their
fraction bits by their corresponding power of two. Since 2-sup-23 +
2-sup-22 + ... + 2-sup-0 is a sum of integers, it must itself be
an integer. Only if the final terms of the sum involve negative
powers of two can it contain fractions.

The other "this" you might be wondering about is: how do you
drop off the fractional bits? *That* one depends (for efficiency
reasons) on the CPU. The two easy ways are bit-twiddling, and
doing addition followed by subtraction. In both cases, we just
want to zero out any mantissa (fraction) bits that represent
negative powers of two. The bit-twiddling method does it with
method uses the normalization hardware to knock them out. If
normalization is slow (e.g., done in software or with a microcode
loop), the bit-twiddling method is generally faster.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
Reading email is like searching for food in the garbage, thanks to spammers.

Chris Torek, Oct 22, 2004
5. ### Gordon BurdittGuest

>I'm using the expression "int a = ceil( SomeDouble )".
>The man page says that ceil returns the smallest
>integer that is not less than SomeDouble, represented
>as a double. However, my understanding is that a
>double has nonuniform precision throughout its value
>range. Will a double always be able to exactly
>represent any value of type int?

No. There is nothing prohibiting an implementation from choosing
int = 64-bit signed integer, and double = 64-bit IEEE double, which
has only 53 mantissa bits. Integers outside the range +/- 2**53
may be rounded.

>point me to an explanation of how this is ensured,
>given that the details of a type realization varies
>with the platform?

It is NOT ensured.

Gordon L. Burditt

Gordon Burditt, Oct 22, 2004
6. ### Erik TrulssonGuest

Fred Ma <> wrote:
> I'm using the expression "int a = ceil( SomeDouble )".
> The man page says that ceil returns the smallest
> integer that is not less than SomeDouble, represented
> as a double. However, my understanding is that a
> double has nonuniform precision throughout its value
> range.

I am not sure what you mean here, but a double is a floating-point type
and like all such has a precision of some fixed number of significant
digits. This precision does not vary, but for large exponents the
difference between one number and the next higher one can be fairly
large.

> Will a double always be able to exactly
> represent any value of type int?

Not necessarily. If, as is common, a double is 64 bits wide with 53
bits of precision, and (as is less common) int is also 64 bits wide
then there are some values of type int which can not be exactly
represented by a double.

> point me to an explanation of how this is ensured,
> given that the details of a type realization varies
> with the platform?
>
> Thanks.
>
> Fred
>
> P.S. I am not worried about overflowing the int value
> range, just about the guaranteed precise representation
> of int by double.

--
Erik Trulsson

Erik Trulsson, Oct 22, 2004
7. ### Erik TrulssonGuest

Fred Ma <> wrote:
> Rouben Rostamian wrote:
>>
>> I don't know whether the C Standard specifies anything to
>> this effect. But here is an implementation-specific
>> observation.
>>
>> On a machine with 64-bit doubles which follow the IEEE
>> specification, the mantissa part is 53 bits (plus one hidden
>> bit as well) therefore integers as large as around 2 to the
>> 50th power should be exactly representable. In particular, if
>> the machine has 32-bit ints, they are all exactly representable
>> as doubles.
>>
>> On my machine, which has 32-bit ints and 64-bit doubles,
>> the following yields the exact answer:
>>
>> printf("%30.15f\n", 1.0 + pow(2.0, 52.));
>>
>> However the following stretches it too far and the answer
>> is inexact:
>>
>> printf("%30.15f\n", 1.0 + pow(2.0, 53.));

>
> I realize that if a double actually uses twice as many bits as
> ints, the mantissa should be big enough that imprecision should
> never arise. I'm just concerned about whether this can be relied
> upon.

This can't be relied upon.

> My faith in what seems normal has been shaken after finding
> that long has the same number of bits as int in some environments.

Actually in most environments these days. (Most Unix-variants on
32-bit systems has both int and as 32 bits wide.)

> What if double has the same number of bits as ints in some
> environments? Some of those bits will be taken up by the
> exponent, and the mantissa will actually have fewer bits than an
> int. Hence, it will be less precise than ints within the value
> range of ints.

Correct, and this can indeed happen.

--
Erik Trulsson

Erik Trulsson, Oct 22, 2004
8. ### Jack KleinGuest

On 22 Oct 2004 00:07:14 GMT, Fred Ma <> wrote in
comp.lang.c:

> I'm using the expression "int a = ceil( SomeDouble )".
> The man page says that ceil returns the smallest
> integer that is not less than SomeDouble, represented
> as a double. However, my understanding is that a
> double has nonuniform precision throughout its value
> range. Will a double always be able to exactly
> represent any value of type int? Could someone please
> point me to an explanation of how this is ensured,
> given that the details of a type realization varies
> with the platform?
>
> Thanks.
>
> Fred
>
> P.S. I am not worried about overflowing the int value
> range, just about the guaranteed precise representation
> of int by double.

As others have mentioned, on 64-bit platforms some integer types, and
perhaps even type int on some, have 64 bits and doubles usually have
fewer mantissa bits than this.

What I haven't seen anyone else point out, so far, is the fact that
this implementation-defined characteristic is available to your
program via the macros DECIMAL_DIG and DBL_DIG in <float.h>.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~ajo/docs/FAQ-acllc.html

Jack Klein, Oct 22, 2004
9. ### Fred MaGuest

Fred Ma wrote:
>
> I'm using the expression "int a = ceil( SomeDouble )". The man page says
> that ceil returns the smallest integer that is not less than SomeDouble,
> represented as a double. However, my understanding is that a double has
> nonuniform precision throughout its value range. Will a double always be
> able to exactly represent any value of type int? Could someone please
> point me to an explanation of how this is ensured, given that the details
> of a type realization varies with the platform?
>
> Thanks.
>
> Fred
>
> P.S. I am not worried about overflowing the int value range, just about
> the guaranteed precise representation of int by double.

Thanks, all, for your replies. They have pointed out a flaw with my own
question. Specifically, it is one thing to ask:

(1) if a double can precisely represent any int.

It is quite another to ask:

(2) if an int(ceil(SomeDouble)) can precisely represent the smallest
integer that is no smaller than SomeDouble, given that SomeDouble is
in the value range of int.

The answer to #1 is clearly no if the mantissa of the double has
"significantly" fewer bits than the int. The reason for "significantly" is
approximate bookkeeping I've walked through; based on Chris's description,
I tried to sanity check this. It starts with the idea that whether a
double can represent any int depends on whether a double can increase in
value by exactly 1 throughout the value range of int. That is, when the
LSB of the mantissa is toggled, does the value of the double change by no
more than 1? For a mantissa of N bits, ignoring the IEEE hidden bit, this
condition is satisfied if scaling due to the exponent (power of 2) is
less-than-or-equal-to 2^N. I'm not talking about how the exponent is
represented in terms of bits; I'm talking about multiplying the mantissa by
2^N, however it is represented in IEEE format. Bascially, the scaling is
such that there are no fractional bits. An exponent value greater than N
yields a scaling that causes the double to increment by more than 1 when
the mantissa increments. Hence, the limiting condition for the double to
have a precision of unity is when the scaling is 2^N. The maximum number
under this condition is when the mantissa is all-ones (N+1 ones including
the hidden bit) i.e. the double has value 2^(N+2)-1. (I'm ignoring the
details to accommodate negative numbers, this might affect the answer by a
bit or so). If all ints fall within this limit, then a double can
represent all ints.

I think the answer to #2 follows from this picture of scaling the mantissa
so that the LSB has unit value. I had to remind myself that the condition
considered in #2 is that SomeDouble is within the value range of int, so
the hazard being tested is not one of overflow. Irrespective of this
condition, however, there are two scenarios which ceil(SomeDouble) can be
split into. One is that the exponent scaling of SomeDouble leaves some
fractional bits, and the other is that it doesn't. If there are some
fractional bits, then the resolution of SomeDouble in that value range is
obviously more precise than a unity step, so integers are precisely
representable, and ceil should return the right value. If there are no
fractional bits, then SomeDouble has an integral value, and passing it
through the ceil function should result in no change, regardless of the
resolution of SomeDouble in that value range i.e. ceil should be able to
return the correct value as a double.

The unintuitive result of this (to me) is that SomeDouble *always* returns
precisely the right answer. Whether it fits into an int is a different
issue (issue#1). I suspect this is what Chris was illustrating.

Fred

Fred Ma, Oct 22, 2004
10. ### Fred MaGuest

Jack Klein wrote:
>
> As others have mentioned, on 64-bit platforms some integer types, and
> perhaps even type int on some, have 64 bits and doubles usually have
> fewer mantissa bits than this.
>
> What I haven't seen anyone else point out, so far, is the fact that
> this implementation-defined characteristic is available to your
> program via the macros DECIMAL_DIG and DBL_DIG in <float.h>.

Hi, Jack,

I found these definitions at Dinkum:

DECIMAL_DIG
#define DECIMAL_DIG <#if expression >= 10> [added with C99]
The macro yields the minimum number of decimal digits needed to represent all the significant digits for type long double.

FLT_DIG
#define FLT_DIG <#if expression >= 6>
The macro yields the precision in decimal digits for type float.

I guess the point is that one can infer the bit-width of the mantissa from
them. Thanks.

Fred

Fred Ma, Oct 22, 2004
11. ### dandelionGuest

"Fred Ma" <> wrote in message
news:...
<snip>
> Will a double always be able to exactly
> represent any value of type int?

Wether (strictly speaking) it will or won't I wouldn't dare to say given the
plethora of representations in use. What I *can* say from my own expirience
is "Do not count on it".

Since the mantissa can (within its limits) represent an integer exactly, you
can simply set the exponent to 1 and the integer could be represented
exactly. However, M_PI/M_PI seldomly equals 1.000000.

dandelion, Oct 22, 2004
12. ### Michael MairGuest

Fred Ma wrote:
> Jack Klein wrote:
>
>>As others have mentioned, on 64-bit platforms some integer types, and
>>perhaps even type int on some, have 64 bits and doubles usually have
>>fewer mantissa bits than this.
>>
>>What I haven't seen anyone else point out, so far, is the fact that
>>this implementation-defined characteristic is available to your
>>program via the macros DECIMAL_DIG and DBL_DIG in <float.h>.

>
> I found these definitions at Dinkum:
>
> DECIMAL_DIG
> #define DECIMAL_DIG <#if expression >= 10> [added with C99]
> The macro yields the minimum number of decimal digits needed to represent all the significant digits for type long double.
>
> FLT_DIG
> #define FLT_DIG <#if expression >= 6>
> The macro yields the precision in decimal digits for type float.
>
> I guess the point is that one can infer the bit-width of the mantissa from
> them. Thanks.

Umh, for the "bit width" rather use DBL_MANT_DIG, after you made
sure that FLT_RADIX is 2 (which is the base you expect).
If you want to know the highest exactly representable number (in the
"contiguous" subset, of course), you can calculate it from there or use
(assuming base 2) 2.0/DBL_EPSILON. Use a conversion to unsigned int and
back to find out whether unsigned can hold this value.

Cheers
Michael
--
E-Mail: Mine is a /at/ gmx /dot/ de address.

Michael Mair, Oct 22, 2004
13. ### Fred MaGuest

dandelion wrote:
>
> M_PI/M_PI seldomly equals 1.000000.

I imagine that would depend on how division is implemented.

Fred

Fred Ma, Oct 22, 2004
14. ### Fred MaGuest

Michael Mair wrote:
>
> Fred Ma wrote:
> > Jack Klein wrote:
> >
> >>As others have mentioned, on 64-bit platforms some integer types, and
> >>perhaps even type int on some, have 64 bits and doubles usually have
> >>fewer mantissa bits than this.
> >>
> >>What I haven't seen anyone else point out, so far, is the fact that
> >>this implementation-defined characteristic is available to your
> >>program via the macros DECIMAL_DIG and DBL_DIG in <float.h>.

> >
> > I found these definitions at Dinkum:
> >
> > DECIMAL_DIG
> > #define DECIMAL_DIG <#if expression >= 10> [added with C99]
> > The macro yields the minimum number of decimal digits needed to represent all the significant digits for type long double.
> >
> > FLT_DIG
> > #define FLT_DIG <#if expression >= 6>
> > The macro yields the precision in decimal digits for type float.
> >
> > I guess the point is that one can infer the bit-width of the mantissa from
> > them. Thanks.

>
> Umh, for the "bit width" rather use DBL_MANT_DIG, after you made
> sure that FLT_RADIX is 2 (which is the base you expect).
> If you want to know the highest exactly representable number (in the
> "contiguous" subset, of course), you can calculate it from there or use
> (assuming base 2) 2.0/DBL_EPSILON. Use a conversion to unsigned int and
> back to find out whether unsigned can hold this value.

Thanks, Michael.

Fred

Fred Ma, Oct 22, 2004
15. ### dandelionGuest

"Fred Ma" <> wrote in message
news:...
> dandelion wrote:
> >
> > M_PI/M_PI seldomly equals 1.000000.

>
> I imagine that would depend on how division is implemented.
>
> Fred

----- Original Message -----
From: "Fred Ma" <>
Newsgroups: comp.lang.c
Sent: Friday, October 22, 2004 2:03 PM
Subject: Re: Can a double always represent an int exactly?

> dandelion wrote:
> >
> > M_PI/M_PI seldomly equals 1.000000.

>
> I imagine that would depend on how division is implemented.

Of course, that's why I wrote "seldomly". And which implementation would
return 1.000000, exactly? I'm curious. Try a few CPU's/FPU's and check the
results. I'll buy you a beer if
you find one.

I wonder why all that 'epsilon-squared' stuff was good for back in HIO and
why the informatics teacher kept hammering us with "Never compare two floats
for equality! Never!".

Must have been a geek, worrying about such detail.

dandelion, Oct 22, 2004
16. ### Keith ThompsonGuest

"dandelion" <> writes:
> "Fred Ma" <> wrote in message
>> dandelion wrote:
>> >
>> > M_PI/M_PI seldomly equals 1.000000.

>>
>> I imagine that would depend on how division is implemented.

>
> Of course, that's why I wrote "seldomly". And which implementation would
> return 1.000000, exactly? I'm curious. Try a few CPU's/FPU's and check the
> results. I'll buy you a beer if
> you find one.

I just tried this on a wide variety of systems; M_PI/M_PI compares
equal to 1.0 on all but one of them. (The exception was a Cray SV1.)

Here's the program I used:

#include <stdio.h>
#include <math.h>
int main(void)
{
double var_M_PI = M_PI;
double ratio = M_PI / M_PI;
double var_ratio = var_M_PI / var_M_PI;
printf("M_PI = %g\n", M_PI);
printf("var_M_PI = %g\n", var_M_PI);
printf("ratio = %g\n", ratio);
printf("ratio %s 1.0\n", ratio == 1.0 ? "==" : "!=");
printf("var_ratio = %g\n", var_ratio);
printf("var_ratio %s 1.0\n", var_ratio == 1.0 ? "==" : "!=");
return 0;
}

Caveats: A moderately clever compiler could compute the value at
compilation time (I didn't check this, but I didn't use any
optimization options). And of course M_PI is non-standard.

--
Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

Keith Thompson, Oct 22, 2004
17. ### Fred MaGuest

Keith Thompson wrote:
>
> "dandelion" <> writes:
> > "Fred Ma" <> wrote in message
> >> dandelion wrote:
> >> >
> >> > M_PI/M_PI seldomly equals 1.000000.
> >>
> >> I imagine that would depend on how division is implemented.

> >
> > Of course, that's why I wrote "seldomly". And which implementation would
> > return 1.000000, exactly? I'm curious. Try a few CPU's/FPU's and check the
> > results. I'll buy you a beer if
> > you find one.

>
> I just tried this on a wide variety of systems; M_PI/M_PI compares
> equal to 1.0 on all but one of them. (The exception was a Cray SV1.)
>
> Here's the program I used:
>
> #include <stdio.h>
> #include <math.h>
> int main(void)
> {
> double var_M_PI = M_PI;
> double ratio = M_PI / M_PI;
> double var_ratio = var_M_PI / var_M_PI;
> printf("M_PI = %g\n", M_PI);
> printf("var_M_PI = %g\n", var_M_PI);
> printf("ratio = %g\n", ratio);
> printf("ratio %s 1.0\n", ratio == 1.0 ? "==" : "!=");
> printf("var_ratio = %g\n", var_ratio);
> printf("var_ratio %s 1.0\n", var_ratio == 1.0 ? "==" : "!=");
> return 0;
> }
>
> Caveats: A moderately clever compiler could compute the value at
> compilation time (I didn't check this, but I didn't use any
> optimization options). And of course M_PI is non-standard.

Seriously, I wasn't implying that practical implementations of
division were necessarily sophisticated enough to recognize
equivalence of numerator and denominator. What I should ahve
said was that I can see such a discrepancy arising, since
division is not straightforward to implement. I'm talking about
cases that aren't optimized away at compile time.

Fred

Fred Ma, Oct 23, 2004
18. ### Chris TorekGuest

A few minor corrections...

In article <> I wrote (in part):
>>.. a double has nonuniform precision throughout its value range.

>
>This is correct (well, I can imagine a weird implementation that
>deliberately makes "double"s have constant precision by often
>wasting a lot of space; it seems quite unlikely though).

It occurs to me now that "precision" is not properly defined here.
When dealing with scientific notation and decimal numbers, something
like 1.23e+10 is less precise than 1.230e+10. The precision here
is determined by the number of digits in the mantissa (which is
why we have to use the "e+10" notation to suppress "unwanted"
trailing zeros).

Using this definition of precision, and keeping in mind that most
computers today use powers of 2 (binary floating point) rather than
powers of ten (decimal floating point), we actually do have "constant
precision", such as "always exactly 24 bits of mantissa" (provided
we ignore those pesky "denorms" ).

This is of course not what the original poster and I meant by
"precision" (as illustrated below) -- we were referring to digits
beyond the decimal point after conversion to printed form via "%f",
for instance. Note, however, that IBM "hex float" (as used on
S/360 -- floating point with a radix of 16 instead of 2) really
*does* have "precision wobble": the number of "useful" bits in the
mantissa changes as numbers change in magnitude. This gives the
numerical analysis folks headaches. IEEE floating point is rather
better behaved.

I need to fix one more typo though:

>... [using] "float" ... on most of today's implementations.
>Here, we get "interesting" effects near 8388608.0 and 16777216.0.
>Values below 16777216.0 step by ones: 8388608.0 is followed
>immediately by 8388609.0, for instance, and 16777215.0 is followed
>immediately by 16777216.0. On the other hand, below (float)(1<<23)
>or above (float)(1<<24), we step by 1/2 or 2 respectively. Using
>nextafterf() (if you have it) and variables set to the right values,
>you might printf() some results and find:
>
> nextafterf(8388608.0, -inf) = 8388607.5
> nextafterf(16777216.0, +inf) = 16777216.2

nextafterf(16777216.0, +inf) = 16777218.0

(I typed this all in manually, rather than writing C code to
call nextafterf(), display the results as above, and then
the change by hand.)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
Reading email is like searching for food in the garbage, thanks to spammers.

Chris Torek, Oct 23, 2004
19. ### Fred MaGuest

Chris Torek wrote:
>
> A few minor corrections...
>
> In article <> I wrote (in part):
> >>.. a double has nonuniform precision throughout its value range.

> >
> >This is correct (well, I can imagine a weird implementation that
> >deliberately makes "double"s have constant precision by often
> >wasting a lot of space; it seems quite unlikely though).

>
> It occurs to me now that "precision" is not properly defined here.
> When dealing with scientific notation and decimal numbers, something
> like 1.23e+10 is less precise than 1.230e+10. The precision here
> is determined by the number of digits in the mantissa (which is
> why we have to use the "e+10" notation to suppress "unwanted"
> trailing zeros).
>
> Using this definition of precision, and keeping in mind that most
> computers today use powers of 2 (binary floating point) rather than
> powers of ten (decimal floating point), we actually do have "constant
> precision", such as "always exactly 24 bits of mantissa" (provided
> we ignore those pesky "denorms" ).
>
> This is of course not what the original poster and I meant by
> "precision" (as illustrated below) -- we were referring to digits
> beyond the decimal point after conversion to printed form via "%f",
> for instance. Note, however, that IBM "hex float" (as used on
> S/360 -- floating point with a radix of 16 instead of 2) really
> *does* have "precision wobble": the number of "useful" bits in the
> mantissa changes as numbers change in magnitude. This gives the
> numerical analysis folks headaches. IEEE floating point is rather
> better behaved.
>
> I need to fix one more typo though:
>
> >... [using] "float" ... on most of today's implementations.
> >Here, we get "interesting" effects near 8388608.0 and 16777216.0.
> >Values below 16777216.0 step by ones: 8388608.0 is followed
> >immediately by 8388609.0, for instance, and 16777215.0 is followed
> >immediately by 16777216.0. On the other hand, below (float)(1<<23)
> >or above (float)(1<<24), we step by 1/2 or 2 respectively. Using
> >nextafterf() (if you have it) and variables set to the right values,
> >you might printf() some results and find:
> >
> > nextafterf(8388608.0, -inf) = 8388607.5
> > nextafterf(16777216.0, +inf) = 16777216.2

>
> This last line should read:
>
> nextafterf(16777216.0, +inf) = 16777218.0
>
> (I typed this all in manually, rather than writing C code to
> call nextafterf(), display the results as above, and then
> the change by hand.)

Chris, thanks for the correction. I think I got the gist of
it from your original post. I did a blanket reply elaborating on it,
Fri. Oct. 22 Message-ID <>. Thanks
for helping me get my brain around it, and if you have any comments
on that, I'm certainly interested.

Fred

Fred Ma, Oct 23, 2004
20. ### Dik T. WinterGuest

In article <> Fred Ma <> writes:
....
> Seriously, I wasn't implying that practical implementations of
> division were necessarily sophisticated enough to recognize
> equivalence of numerator and denominator. What I should ahve
> said was that I can see such a discrepancy arising, since
> division is not straightforward to implement. I'm talking about
> cases that aren't optimized away at compile time.

It is not straightforward to implement. Nevertheless, whenever the FPU
conforms to the IEEE standard the division *must* deliver the exact
answer if the quotient is representable. So on all systems using such
FPU's (and that is the majority at this moment) should deliver 1.0 when
confronted with a/a, in whatever way it is disguised. To get division
right is not straigthforward, but it is not so very difficult either.

That Keith Thompson found that it was not the case on a Cray SV1 is
entirely because that system has not an IEEE conforming floating point
system. (That machine does not have a divide instruction. It
calculates an approximation of the inverse of the denominator and
multiplies with the numerator, and one Newton iteration is performed.
Due to some quirks it may give an inexact result. If I remember
right, the smallest integral division that is inexact is 17.0/17.0.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

Dik T. Winter, Oct 24, 2004