# converting float to double

Discussion in 'C Programming' started by Dilip, Dec 19, 2006.

1. ### DilipGuest

Recently in our code, I ran into a situation where were stuffing a
float inside a double. The precision was extended automatically
because of that. To make a long story short, this caused problems
elsewhere in another part of the system where that figure was used for
some calculation and some eventual truncation led to the system going
haywire. So my question is, given this code:

int main()
{
float f = 59.89F;
/* the following line caused problems */
double d = f;

/* they ended up solving it like this */
f *= 1000.0;
double dd = (double)f / 1000.0;

return 0;
}

I see what is being done but why does the latter make the situation
better?
(consider 'f' in real life to hold stock prices)

Dilip, Dec 19, 2006

2. ### Guest

Dilip wrote:
> Recently in our code, I ran into a situation where were stuffing a
> float inside a double. The precision was extended automatically
> because of that. To make a long story short, this caused problems
> elsewhere in another part of the system where that figure was used for
> some calculation and some eventual truncation led to the system going
> haywire. So my question is, given this code:
>
> int main()
> {
> float f = 59.89F;
> /* the following line caused problems */
> double d = f;
>
> /* they ended up solving it like this */
> f *= 1000.0;
> double dd = (double)f / 1000.0;
>
> return 0;
> }
>
> I see what is being done but why does the latter make the situation
> better?
> (consider 'f' in real life to hold stock prices)

Not a good idea...

Floating point is by its nature imprecise - multiplying by 1000 simply
adjusts the imprecision in a way which was helpful in this specific
case, but may not be helpful in all cases (AFAIK).

If you want to safely and accurately handle this sort of data, sources
such as http://www2.hursley.ibm.com/decimal/ may be helpful.

, Dec 19, 2006

3. ### CBFalconerGuest

Dilip wrote:
>
> Recently in our code, I ran into a situation where were stuffing a
> float inside a double. The precision was extended automatically
> because of that. To make a long story short, this caused problems
> elsewhere in another part of the system where that figure was used
> for some calculation and some eventual truncation led to the system
> going haywire. So my question is, given this code:
>
> int main()
> {
> float f = 59.89F;
> /* the following line caused problems */
> double d = f;
>
> /* they ended up solving it like this */
> f *= 1000.0;
> double dd = (double)f / 1000.0;
>
> return 0;
> }
>
> I see what is being done but why does the latter make the
> situation better?
> (consider 'f' in real life to hold stock prices)

It doesn't. It should make it dead wrong. Either you are not
showing the real situation or you have a compiler bug.

Alternatively, if you are depending on exact values from floats
and/or doubles, your code is broken. Since stock prices (today)
are decimalized, just use a normalized integer of suitable size.
Read Knuth (TAOCP) on floating point and expected errors.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

CBFalconer, Dec 19, 2006
4. ### peteGuest

Dilip wrote:
>
> Recently in our code, I ran into a situation where were stuffing a
> float inside a double. The precision was extended automatically
> because of that. To make a long story short, this caused problems
> elsewhere in another part of the system where that figure was used for
> some calculation and some eventual truncation led to the system going
> haywire. So my question is, given this code:
>
> int main()
> {
> float f = 59.89F;
> /* the following line caused problems */
> double d = f;
>
> /* they ended up solving it like this */
> f *= 1000.0;
> double dd = (double)f / 1000.0;
>
> return 0;
> }
>
> I see what is being done but why does the latter make the situation
> better?
> (consider 'f' in real life to hold stock prices)

I wouldn't have used a float variable like that to begin with.

Use float,
when you need the smallest floating point type.

Use long double,
when you need the floating point type
with most precision and or range.

Use double,
all the rest of the time.

--
pete

pete, Dec 19, 2006
5. ### DilipGuest

CBFalconer wrote:
> Dilip wrote:
> >
> > Recently in our code, I ran into a situation where were stuffing a
> > float inside a double. The precision was extended automatically
> > because of that. To make a long story short, this caused problems
> > elsewhere in another part of the system where that figure was used
> > for some calculation and some eventual truncation led to the system
> > going haywire. So my question is, given this code:
> >
> > int main()
> > {
> > float f = 59.89F;
> > /* the following line caused problems */
> > double d = f;
> >
> > /* they ended up solving it like this */
> > f *= 1000.0;
> > double dd = (double)f / 1000.0;
> >
> > return 0;
> > }
> >
> > I see what is being done but why does the latter make the
> > situation better?
> > (consider 'f' in real life to hold stock prices)

>
> It doesn't. It should make it dead wrong. Either you are not
> showing the real situation or you have a compiler bug.

I am showing exactly what is being done in our codebase. Why do you
say there is a compiler bug? If you try the example, as soon as the
1st line is executed the variable f contains 59.889999. After its
stuffed into double d, it becomes 59.889999389648437 (I know these
values vary after a certain decimal position). The
multiplication/division to 1000.0 however produces 59.890000000000001
in dd. The management somehow feels thats a more "accurate"
representation. I guess after rounding off to 2 decimal places they
expected 59.89 but a bug somewhere in the code ensured that the former
case got rounded off to 59.88. It was off by a penny and that
triggered major problems in other calculations.

> Since stock prices (today)
> are decimalized, just use a normalized integer of suitable size.
> Read Knuth (TAOCP) on floating point and expected errors.

I am not exactly unfamiliar with floating point representations and its
associated accuracy related problems but I think this is the first time
I am working in a domain where such problems crop up on a more frequent
basis. Could you please elaborate on that "normalized integer of
suitable size"?

Dilip, Dec 19, 2006
6. ### Keith ThompsonGuest

CBFalconer <> writes:
> Dilip wrote:
>> Recently in our code, I ran into a situation where were stuffing a
>> float inside a double. The precision was extended automatically
>> because of that. To make a long story short, this caused problems
>> elsewhere in another part of the system where that figure was used
>> for some calculation and some eventual truncation led to the system
>> going haywire. So my question is, given this code:
>>
>> int main()
>> {
>> float f = 59.89F;
>> /* the following line caused problems */
>> double d = f;
>>
>> /* they ended up solving it like this */
>> f *= 1000.0;
>> double dd = (double)f / 1000.0;
>>
>> return 0;
>> }
>>
>> I see what is being done but why does the latter make the
>> situation better?
>> (consider 'f' in real life to hold stock prices)

>
> It doesn't. It should make it dead wrong. Either you are not
> showing the real situation or you have a compiler bug.

Are you sure about that? The code multiplies f by 1000.0, then
divides the result by 1000.0. The *mathematical* result is the same,
but the way it's done is likely to change the rounding behavior
slightly. It's not likely to be a general solution to any problem.
It smacks of "cargo cult programming"; see
<http://www.catb.org/~esr/jargon/html/C/cargo-cult-programming.html>.

> Alternatively, if you are depending on exact values from floats
> and/or doubles, your code is broken. Since stock prices (today)
> are decimalized, just use a normalized integer of suitable size.
> Read Knuth (TAOCP) on floating point and expected errors.

Another good resource is Goldberg's paper "What Every Computer
floating" to find it.

--
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, Dec 19, 2006
7. ### William HughesGuest

pete wrote:
> Dilip wrote:
> >
> > Recently in our code, I ran into a situation where were stuffing a
> > float inside a double. The precision was extended automatically
> > because of that. To make a long story short, this caused problems
> > elsewhere in another part of the system where that figure was used for
> > some calculation and some eventual truncation led to the system going
> > haywire. So my question is, given this code:
> >
> > int main()
> > {
> > float f = 59.89F;
> > /* the following line caused problems */
> > double d = f;
> >
> > /* they ended up solving it like this */
> > f *= 1000.0;
> > double dd = (double)f / 1000.0;
> >
> > return 0;
> > }
> >
> > I see what is being done but why does the latter make the situation
> > better?
> > (consider 'f' in real life to hold stock prices)

>
> I wouldn't have used a float variable like that to begin with.
>
> Use float,
> when you need the smallest floating point type.
>
> Use long double,
> when you need the floating point type
> with most precision and or range.
>
> Use double,
> all the rest of the time.
>

This comes squarely under the heading "If you can understand this

- William Hughes
> --
> pete

William Hughes, Dec 19, 2006
8. ### Random832Guest

2006-12-19 <>,
pete wrote:
> Dilip wrote:
>>
>> Recently in our code, I ran into a situation where were stuffing a
>> float inside a double. The precision was extended automatically
>> because of that. To make a long story short, this caused problems
>> elsewhere in another part of the system where that figure was used for
>> some calculation and some eventual truncation led to the system going
>> haywire. So my question is, given this code:
>>
>> int main()
>> {
>> float f = 59.89F;
>> /* the following line caused problems */
>> double d = f;
>>
>> /* they ended up solving it like this */
>> f *= 1000.0;
>> double dd = (double)f / 1000.0;
>>
>> return 0;
>> }
>>
>> I see what is being done but why does the latter make the situation
>> better?
>> (consider 'f' in real life to hold stock prices)

>
> I wouldn't have used a float variable like that to begin with.
>
> Use float,
> when you need the smallest floating point type.
>
> Use long double,
> when you need the floating point type
> with most precision and or range.

Use integers,
when you need to be exact to within a known specific unit, such as
a dollar, cent, or Nth part of either.

> Use double,
> all the rest of the time.

Random832, Dec 19, 2006
9. ### Random832Guest

2006-12-19 <>,
Dilip wrote:
> I am not exactly unfamiliar with floating point representations and its
> associated accuracy related problems but I think this is the first time
> I am working in a domain where such problems crop up on a more frequent
> basis. Could you please elaborate on that "normalized integer of
> suitable size"?

Have an integer variable [probably type long or long long] counting the
number of cents (or thousandths of a dollar, or whatever). Doing your
weird cast thing fixed the error in that one particular case, but could
have introduced other errors elsewhere.

Random832, Dec 19, 2006
10. ### DilipGuest

Keith Thompson wrote:
> CBFalconer <> writes:
> > Dilip wrote:
> >> Recently in our code, I ran into a situation where were stuffing a
> >> float inside a double. The precision was extended automatically
> >> because of that. To make a long story short, this caused problems
> >> elsewhere in another part of the system where that figure was used
> >> for some calculation and some eventual truncation led to the system
> >> going haywire. So my question is, given this code:
> >>
> >> int main()
> >> {
> >> float f = 59.89F;
> >> /* the following line caused problems */
> >> double d = f;
> >>
> >> /* they ended up solving it like this */
> >> f *= 1000.0;
> >> double dd = (double)f / 1000.0;
> >>
> >> return 0;
> >> }
> >>
> >> I see what is being done but why does the latter make the
> >> situation better?
> >> (consider 'f' in real life to hold stock prices)

> >
> > It doesn't. It should make it dead wrong. Either you are not
> > showing the real situation or you have a compiler bug.

>
> Are you sure about that? The code multiplies f by 1000.0, then
> divides the result by 1000.0. The *mathematical* result is the same,
> but the way it's done is likely to change the rounding behavior
> slightly. It's not likely to be a general solution to any problem.
> It smacks of "cargo cult programming"; see
> <http://www.catb.org/~esr/jargon/html/C/cargo-cult-programming.html>.

I totally agree which is the reason why I posted to this NG to
understand what exactly is being solved by that 1000.0 workaround.

> Another good resource is Goldberg's paper "What Every Computer
> floating" to find it.

I came across that paper quite a few times in the past. I never
realized one day I would need to know that amount of detail. Do you
have any insights into that normalized integer thing everyone has been
pointing out so far? If one of you can give me an example, i'd
appreciate it.

Dilip, Dec 19, 2006
11. ### Guest

Dilip wrote:
> I came across that paper quite a few times in the past. I never
> realized one day I would need to know that amount of detail. Do you
> have any insights into that normalized integer thing everyone has been
> pointing out so far? If one of you can give me an example, i'd
> appreciate it.

Store the amount as pennies in a long long integer. The Goldberg paper
is a must read if you are doing numerics with other people's money.
IMO-YMMV.

#include <stdio.h>
#include <stdlib.h>

char string[256];
int main(void)
{
long long pennies;

try_again:
puts("Enter a cost in pennies:");
if (fgets(string, sizeof string, stdin)) {
sscanf(string, "%lld", &pennies);
printf("That's %lld dollars and %lld cents or \$%lld.%lld\n",
pennies / 100, pennies % 100, pennies / 100, pennies % 100);
} else
goto try_again;

return EXIT_SUCCESS;
}
/* e.g.:
Enter a cost in pennies:
99999999
That's 999999 dollars and 99 cents or \$999999.99
*/

, Dec 19, 2006
12. ### MalcolmGuest

"Dilip" <> wrote in message
news:...
>
> Recently in our code, I ran into a situation where were stuffing a
> float inside a double. The precision was extended automatically
> because of that. To make a long story short, this caused problems
> elsewhere in another part of the system where that figure was used for
> some calculation and some eventual truncation led to the system going
> haywire. So my question is, given this code:
>
> int main()
> {
> float f = 59.89F;
> /* the following line caused problems */
> double d = f;
>
> /* they ended up solving it like this */
> f *= 1000.0;
> double dd = (double)f / 1000.0;
>
> return 0;
> }
>
> I see what is being done but why does the latter make the situation
> better?
> (consider 'f' in real life to hold stock prices)
>

That is called hacking.
Somehow a cent must have been knocked off in the code - decimals cannot be
represented perfectly in binary floating point notation - leading the
program to suspect salami tactics. Multiplying by 1000 converts the number
to an integer, and the division is done in double precision, so the cent
might not have been dropped.

This approach to problem solving, incidentally, is known as hacking, and is
an indication that the company's management is in a serious mess.
--
www.personal.leeds.ac.uk/~bgy1mm

Malcolm, Dec 19, 2006
13. ### CBFalconerGuest

Dilip wrote:
> Keith Thompson wrote:
>> CBFalconer <> writes:
>>> Dilip wrote:
>>>
>>>> Recently in our code, I ran into a situation where were stuffing a
>>>> float inside a double. The precision was extended automatically
>>>> because of that. To make a long story short, this caused problems
>>>> elsewhere in another part of the system where that figure was used
>>>> for some calculation and some eventual truncation led to the system
>>>> going haywire. So my question is, given this code:
>>>>
>>>> int main()
>>>> {
>>>> float f = 59.89F;
>>>> /* the following line caused problems */
>>>> double d = f;
>>>>
>>>> /* they ended up solving it like this */
>>>> f *= 1000.0;
>>>> double dd = (double)f / 1000.0;
>>>>
>>>> return 0;
>>>> }
>>>>
>>>> I see what is being done but why does the latter make the
>>>> situation better?
>>>> (consider 'f' in real life to hold stock prices)
>>>
>>> It doesn't. It should make it dead wrong. Either you are not
>>> showing the real situation or you have a compiler bug.

>>
>> Are you sure about that? The code multiplies f by 1000.0, then
>> divides the result by 1000.0. The *mathematical* result is the same,
>> but the way it's done is likely to change the rounding behavior
>> slightly. It's not likely to be a general solution to any problem.
>> It smacks of "cargo cult programming"; see
>> <http://www.catb.org/~esr/jargon/html/C/cargo-cult-programming.html>.

>
> I totally agree which is the reason why I posted to this NG to
> understand what exactly is being solved by that 1000.0 workaround.
>
>> Another good resource is Goldberg's paper "What Every Computer
>> floating" to find it.

>
> I came across that paper quite a few times in the past. I never
> realized one day I would need to know that amount of detail. Do you
> have any insights into that normalized integer thing everyone has been
> pointing out so far? If one of you can give me an example, i'd
> appreciate it.

As a temporary measure, try converting the value to a long:

long longvalueincents;

longvalueincents = (f * 100) + 0.5; /* control rounding */

and, to print it out, try:

printf("%ld.%ld", longvalueincents/100, longvalueincents%100);
fflush(stdout);

Note that you don't need any casts. A cast is probably an error.

You can probably incorporate all this in a single function that
dumps the double rounded to two decimals, but reading the printf
documentation is likely to be more rewarding.

You could also hire me. See below.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

CBFalconer, Dec 19, 2006
14. ### MalcolmGuest

<> wrote in message
news:...
> Dilip wrote:
>> I came across that paper quite a few times in the past. I never
>> realized one day I would need to know that amount of detail. Do you
>> have any insights into that normalized integer thing everyone has been
>> pointing out so far? If one of you can give me an example, i'd
>> appreciate it.

>
> Store the amount as pennies in a long long integer. The Goldberg paper
> is a must read if you are doing numerics with other people's money.
> IMO-YMMV.
>

Or in a double, but as an integral number of pence.
That way the maths is accurate up to 48 bits or so, and degrades garcefully
under hyperinflation.
--
www.personal.leeds.ac.uk/~bgy1mm

Malcolm, Dec 20, 2006
15. ### William HughesGuest

Dilip wrote:
> Recently in our code, I ran into a situation where were stuffing a
> float inside a double. The precision was extended automatically
> because of that. To make a long story short, this caused problems
> elsewhere in another part of the system where that figure was used for
> some calculation and some eventual truncation led to the system going
> haywire. So my question is, given this code:
>
> int main()
> {
> float f = 59.89F;
> /* the following line caused problems */
> double d = f;
>
> /* they ended up solving it like this */
> f *= 1000.0;
> double dd = (double)f / 1000.0;
>
> return 0;
> }
>
> I see what is being done but why does the latter make the situation
> better?
> (consider 'f' in real life to hold stock prices)

For the purposes of this explanation treat double as exact
and float as approximate.

The problem is that most exact decimal fractions cannot
be represented exactly by floats. So you need to use
a value that is just a little above or just a little below.

Suppose we have

f=0.0001

then f will have a value slightly below 0.0001
(or maybe not, but it did on my machine just now).

Now multiply by 1000
Now f will have to approximate 0.1. The best approximation is
a value slightly above 0.1 (at least on my machine just now)
So after

1000*f

we get a value slightly above 0.1. We now assign to a double

dd=1000*f

(This is done by first calculating 1000*f as a float and then
assigning to double). We next divide dd by 1000.
dd now has a value slightly greater than 0.0001.

So the net effect of multiplying by 1000, storing in a double, then
dividing by 1000, is to convert a value slightly below 0.0001 to a
value
slightly above 0.0001. Because the code is clearly incompetently
written, it will later blindly truncate, converting a miniscule
difference into
a noticable one.

Of course if you start out with some other value for f, something
different
may happen. Something different may happen on different machines,
or with different compilers, or with different compiler options, or on
different days, or on different runs, or at different times within the
same run.

I wouldn't trust this code in a rigged demo.

- William Hughes.

William Hughes, Dec 20, 2006
16. ### Dik T. WinterGuest

In article <> "Dilip" <> writes:
> Recently in our code, I ran into a situation where were stuffing a
> float inside a double. The precision was extended automatically
> because of that. To make a long story short, this caused problems
> elsewhere in another part of the system where that figure was used for
> some calculation and some eventual truncation led to the system going
> haywire. So my question is, given this code:

....
> float f = 59.89F;
> /* the following line caused problems */
> double d = f;

d will be approximately equal to 59.89 with the same precision as f

> /* they ended up solving it like this */
> f *= 1000.0;

By sheer luck (due to the rounding rules) f is exactly equal to 59890.
(You could even use 100 here with exactly the same results.)

> double dd = (double)f / 1000.0;

And so this is a better approximation in double precision of 59.89.

> I see what is being done but why does the latter make the situation
> better?

It is only luck. For other values of f it could fail and dd could
(I think) even be worse than d.

> (consider 'f' in real life to hold stock prices)

Consider not to use floating point for stock prices. Floating point
inherently carries imprecision (especially if you want to do decimal
calculations), which you do not want with stock prices.
--
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, Dec 20, 2006
17. ### Dik T. WinterGuest

In article <> "Dilip" <> writes:
....
> I am showing exactly what is being done in our codebase. Why do you
> say there is a compiler bug?

There is indeed none, but you are misunderstanding floating point.

> If you try the example, as soon as the
> 1st line is executed the variable f contains 59.889999. After its
> stuffed into double d, it becomes 59.889999389648437 (I know these
> values vary after a certain decimal position).

They do not vary, but you should print with sufficient precision.
The exact value stored in both f and d is:
59.8899993896484375

> The
> multiplication/division to 1000.0 however produces 59.890000000000001
> in dd.

Actually the value stored in dd is:
59.8900000000000005684341886080801486968994140625
But as I stated in another article, it is only luck that this value is
larger than 59.89.

> The management somehow feels thats a more "accurate"
> representation. I guess after rounding off to 2 decimal places they
> expected 59.89 but a bug somewhere in the code ensured that the former
> case got rounded off to 59.88.

Probably not rounding was done but truncation. In that case
when d * 100 is truncated to int it will indeed give 59.88 and
dd * 100 is truncated to 59.89. I can come up with initial values
where this is all reversed. And it is even platform dependent.
See message <>
in comp.arch.arithmetic and the associated thread.

> It was off by a penny and that
> triggered major problems in other calculations.

If off by a penny triggers major problems, you should *really* consider
using integer arithmetic in pennies.

> > Since stock prices (today)
> > are decimalized, just use a normalized integer of suitable size.
> > Read Knuth (TAOCP) on floating point and expected errors.

>
> I am not exactly unfamiliar with floating point representations and its
> associated accuracy related problems but I think this is the first time
> I am working in a domain where such problems crop up on a more frequent
> basis. Could you please elaborate on that "normalized integer of
> suitable size"?

Use 64 bit integers for the amounts in pennies. When you use floating
point and store the amount in dollars you should not be surprised that
after a number of calculations the result is off by a penny (or perhaps
more). Floating point variables can not store amounts in dollars with
precision upto pennies.
--
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, Dec 20, 2006

wrote:

> printf("That's %lld dollars and %lld cents or \$%lld.%lld\n",
> pennies / 100, pennies % 100, pennies / 100, pennies % 100);

printf("That's %lld dollars and %lld cents or \$%lld.%02lld\n",
pennies / 100, pennies % 100, pennies / 100, pennies % 100);

--

19. ### Keith ThompsonGuest

"Malcolm" <> writes:
> <> wrote in message
> news:...
>> Dilip wrote:
>>> I came across that paper quite a few times in the past. I never
>>> realized one day I would need to know that amount of detail. Do you
>>> have any insights into that normalized integer thing everyone has been
>>> pointing out so far? If one of you can give me an example, i'd
>>> appreciate it.

>>
>> Store the amount as pennies in a long long integer. The Goldberg paper
>> is a must read if you are doing numerics with other people's money.
>> IMO-YMMV.
>>

> Or in a double, but as an integral number of pence.
> That way the maths is accurate up to 48 bits or so, and degrades garcefully
> under hyperinflation.

For certain values of "gracefully".

In some contexts, I would think that a one-penny error in a
multi-trillion pound calculation ("trillion" meaning 10**12) would be
considered unacceptable.

64-bit integers should be more than enough to handle anything short of
really extreme hyperinflation with no errors -- assuming that
addition, subtraction, and multiplication by whole numbers are the
only required operations.

Compound interest introduces some interesting problems, but I think
there are very specific rules for how to do those calculations. I
suspect floating-point would not implement those rules correctly.

Warning: I'm musing about stuff I've heard about here and there, but
in which I have no particular expertise.

--
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, Dec 20, 2006
20. ### Keith ThompsonGuest

"Dik T. Winter" <> writes:
[...]
> Consider not to use floating point for stock prices. Floating point
> inherently carries imprecision (especially if you want to do decimal
> calculations), which you do not want with stock prices.

If you need to do calculations with stock prices, find out what the
formal rules are for doing those calculations. (I have no idea what
those rules are, or where you'd find out about them.) Once you've
done that, implement those rules. ("How?" "Correctly.")

--
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, Dec 20, 2006