Rounding Float in C and Remove those Zeros

K

kennethlou

Hi,

If in C a variable appears like X=10.000000, I can round it to zero
decimal places. ie X=10?

I then have to save the number into a variable.

The method appears below:
**************************************
printf("Round % .f \n", 10.0000000);

and the result will appear on the screen as "10"

is there any method to "save" the rounded value as X?

Thanks,
 
V

Vladimir Oka

Hi,
Hi!

If in C a variable appears like X=10.000000, I can round it to zero
decimal places. ie X=10?

Er, yes... maybe. But you don't say what `X` is: an `int`, a `double`?
What exactly do you want to achieve?
I then have to save the number into a variable.

What variable?
The method appears below:
**************************************

This is not "saving a number into a variable".
and the result will appear on the screen as "10"

Yes (not necessarily on the "screen", though).
is there any method to "save" the rounded value as X?

Yes, but you'll have to be more specific about what exactly you want to
do.
 
V

Vladimir Oka

Vladimir Oka opined:
Er, yes... maybe. But you don't say what `X` is: an `int`, a
`double`? What exactly do you want to achieve?


What variable?


This is not "saving a number into a variable".


Yes (not necessarily on the "screen", though).


Yes, but you'll have to be more specific about what exactly you want
to do.

I have also received this reply from Ken:

Firstly, it is considered polite to keep all topical discussions
public. Many people don't even publish their (correct) e-mail
addresses.

In response to your question, to assign 10 to `Y`, you can do either
of:

Y = 10; /* facetious */

or

X = 10.0000;
Y = X;

In the latter, `Y` will indeed be 10 only if 10 is exactly
representable as a floating point number on your implementation. I
think (before looking into the Standard) that all small integers are
guaranteed to be. Beware numbers like 1000000000000042.0000, though.

--
Persistence in one opinion has never been considered
a merit in political leaders.
-- Marcus Tullius Cicero, "Ad familiares", 1st century BC

<http://clc-wiki.net/wiki/Introduction_to_comp.lang.c>
 
O

osmium

Vladimir Oka said:
I > think (before looking into the Standard) that all small integers are
guaranteed to be. Beware numbers like 1000000000000042.0000, though.

No. For example, the number 1 can not be represented exactly. Select as
many terms as you wish from the following series:
1/2, 1/4, 1/8, ....1/2^googolplex

No way are you going to get a 1 out of that. People who write I/O libraries
are aware of this and create the *illusion* that you can.
 
V

Vladimir Oka

osmium opined:
No. For example, the number 1 can not be represented exactly. Select
as many terms as you wish from the following series:
1/2, 1/4, 1/8, ....1/2^googolplex

No way are you going to get a 1 out of that. People who write I/O
libraries are aware of this and create the *illusion* that you can.

Hmmm. A quick look into the Standard, and some searching through past
c.l.c posts seem to justify my statement that certain range of
integers is indeed exactly representable in floating point types.

Look at the <float.h>. It defines FLT_DIG, DBL_DIG and LDBL_DIG as
the number of decimal digits of precision in a `float`, `double` and
`long double`, respectively. Integers with less than this number of
digits should be exactly representable in the appropriate floating
point types.

I can't say much (anything?) about how this is exactly achieved,
though.
 
S

Skarmander

osmium said:
No. For example, the number 1 can not be represented exactly.

You've happened to pick a number that's exactly representable in all
floating-point systems: 1.0 = 1 * b^0 for any valid base b.
Select as many terms as you wish from the following series:
1/2, 1/4, 1/8, ....1/2^googolplex

No way are you going to get a 1 out of that. People who write I/O libraries
are aware of this and create the *illusion* that you can.
This illusion is called rounding, and it's not really I/O specific. It
depends on the number of digits you ask for, and are able to get.

Had you said 0.1, you might have had a point, as this number is not exactly
representable in binary floating point. On the other hand, implementations
need not use base 2; 0.1 *is* exactly representable in base 10.

You may wish to verify that "all small integers" are indeed exactly
representable in every floating-point system. All integers from -(b^p)
through b^p, with b the base and p the precision, are exactly representable.
For b = 2 and p = 24 (an IEEE 754 single-precision float), all integers
from -16777216 through 16777216 are exactly representable. For double
precision, all integers from -(2^53) through 2^53 are exactly representable.

At this point I'm obliged to refer to the classic: "What Every Computer
Scientist Should Know About Floating-Point Arithmetic", available at, for
example,
http://www.physics.ohio-state.edu/~dws/grouplinks/floating_point_math.pdf

S.
 
D

Dik T. Winter

> No. For example, the number 1 can not be represented exactly. Select as
> many terms as you wish from the following series:
> 1/2, 1/4, 1/8, ....1/2^googolplex

You clearly have no idea how floating point works.
 
M

Malcolm

Vladimir Oka said:
Look at the <float.h>. It defines FLT_DIG, DBL_DIG and LDBL_DIG as
the number of decimal digits of precision in a `float`, `double` and
`long double`, respectively. Integers with less than this number of
digits should be exactly representable in the appropriate floating
point types.

I can't say much (anything?) about how this is exactly achieved,
though.
Imagine that you are allowed to represent floating point digits in ASCII
using only six digits.

So a half is no problem. That's "0.5". A small whole number is also no
problem, that's "100" and so forth.
A third is a bit of a snag. We can write "0.3333" which is good enough for
most purposes, but not exact.
Also if we want very large numbers, we run out of digits. However we can go
to scientific notation. So 1234567890 would be "1.23e9". We've lost the low
digits, but we're still accurate to 3 significant figures.

Computer floating point works in a very similar way, except that the digits
are binary rather than decimal, and there isn't a literal point character -
the point is implicit in the scientific style of notation.
 
J

Joe Wright

osmium said:
No. For example, the number 1 can not be represented exactly. Select as
many terms as you wish from the following series:
1/2, 1/4, 1/8, ....1/2^googolplex

No way are you going to get a 1 out of that. People who write I/O libraries
are aware of this and create the *illusion* that you can.
You misunderstand the floating point system. In IEEE 754 integers up to
the width of the mantissa are represented exactly. On my system thats
2**24 for float and 2**53 for double.
 
O

osmium

Joe Wright said:
osmium wrote:
You misunderstand the floating point system. In IEEE 754 integers up to
the width of the mantissa are represented exactly. On my system thats
2**24 for float and 2**53 for double.

You're right. I was referring to the floating point representation I
learned in school. Which was not THE floating point system, it was only A
floating point system. Since seeing some of the earlier posts I have looked
into IEEE 754 and it *is* quite different - and an improvement. Bystanders
should be aware, however, that there is no assurance that C uses IEEE 754,
so coders still have to follow the old rules if they want to permit cross
platform use. That is, don't use == if you want portability. The
representation I was speaking about is still permissible.
 
K

Keith Thompson

osmium said:
You're right. I was referring to the floating point representation I
learned in school. Which was not THE floating point system, it was only A
floating point system. Since seeing some of the earlier posts I have looked
into IEEE 754 and it *is* quite different - and an improvement. Bystanders
should be aware, however, that there is no assurance that C uses IEEE 754,
so coders still have to follow the old rules if they want to permit cross
platform use. That is, don't use == if you want portability. The
representation I was speaking about is still permissible.

And how does this system you learned in school represent 2.0?
 
S

Skarmander

osmium said:
You're right. I was referring to the floating point representation I
learned in school. Which was not THE floating point system, it was only A
floating point system. Since seeing some of the earlier posts I have looked
into IEEE 754 and it *is* quite different - and an improvement. Bystanders
should be aware, however, that there is no assurance that C uses IEEE 754,
so coders still have to follow the old rules if they want to permit cross
platform use. That is, don't use == if you want portability. The
representation I was speaking about is still permissible.
No, it's not. Although a C implementation is not required to use IEEE 754,
it cannot use a system where 1 is not exactly representable as a
floating-point number. Such a system could not meet the requirements of the
floating-point model described in the standard. (As long as we're talking
C99; I don't know what C89 said.) I verified this with the regulars of
comp.std.c to be sure; the thread is at
http://groups.google.com/group/comp.std.c/browse_thread/thread/2389fc9c95445914/dea116e9c14faab5

The "don't use ==" exhortation applies to all floating-point systems, IEEE
754 included. This is not a portability issue, since it's not even
guaranteed (or rather supposed) to work on one platform. If you're talking
binary representations, then those are indeed not portable, but == has
nothing to do with that.

You're sure you're not confusing all this with fixed-point arithmetic? In
fixed-point it's possible for 1.0 not to be exactly representable. (Standard
C does not define any fixed-point types, but it's not hard to implement them.)

S.
 
J

John Bode

osmium said:
No. For example, the number 1 can not be represented exactly. Select as
many terms as you wish from the following series:
1/2, 1/4, 1/8, ....1/2^googolplex

True that the value 1.0 cannot be represented exactly as a sum of the
series 1/2^n (n = -1, -2, -3, ...) with a finite number of terms.
However, since most floating point formats I'm aware of also use an
exponent, it *can* be represented exactly as

0.5 * 2^1

If you're talking about 0.1, now *that's* a problem, even with the
exponent.
 

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