# Why different from expectation

Discussion in 'C Programming' started by yezi, Nov 28, 2005.

1. ### yeziGuest

Hi: All:

I have question with the data type problem. According to the C
standard. IEEE 754

float 4 3.4x10-38E..3.4x10+38E
double 8 1.7x10-308E..1.7x10+308E
long double 12 ???

The code function is (0.000200-0.000300)*(0.000200-0.000300)
suppose the result (difference ) should be the 10 (-8)

difference I claim is double;

But THe code result is :
fp1 0.000200
------------------------
fp2 0.000300
difference is 0.000000

which means the result is zero.

Why?

Thanks
bin YE

yezi, Nov 28, 2005

2. ### Keith ThompsonGuest

"yezi" <> writes:
> I have question with the data type problem. According to the C
> standard. IEEE 754
>
> float 4 3.4x10-38E..3.4x10+38E
> double 8 1.7x10-308E..1.7x10+308E
> long double 12 ???

The C standard allows, but does not require, IEEE 754 floating-point.

> The code function is (0.000200-0.000300)*(0.000200-0.000300)
> suppose the result (difference ) should be the 10 (-8)
>
> difference I claim is double;
>
> But THe code result is :
> fp1 0.000200
> ------------------------
> fp2 0.000300
> difference is 0.000000
>
> which means the result is zero.

It's difficult to tell what problem you're describing. Try posting
some actual code along with the output it produces, and describe how
it differs from the output you expected.

A couple of thoughts:

Floating-point is inexact. For example, the value 0.1 cannot be
represented exactly in binary floating-point; the closest equivalent
in type double, on one system I've tried, is about
0.10000000000000000555111512312578.

If you're using a "%f" output format, it's not going to show very
small values very well. A printed result of 0.000000 doesn't imply
that the value is actually equal to 0.0, just that it's within a
certain range close to 0.0 (which 1.0e-8 probably is). Try the "%g"
format instead; it switches to scientific notation for very large or
very small values.

--
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, Nov 28, 2005

3. ### Keith ThompsonGuest

"yezi" <> writes:
> I have question with the data type problem. According to the C
> standard. IEEE 754
>
> float 4 3.4x10-38E..3.4x10+38E
> double 8 1.7x10-308E..1.7x10+308E
> long double 12 ???
>
> The code function is (0.000200-0.000300)*(0.000200-0.000300)
> suppose the result (difference ) should be the 10 (-8)
>
> difference I claim is double;
>
> But THe code result is :
> fp1 0.000200
> ------------------------
> fp2 0.000300
> difference is 0.000000
>
> which means the result is zero.

One more thing I forgot to mention: Read section 14 of the C FAQ.
<http://www.eskimo.com/~scs/C-faq/top.html>.

--
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, Nov 28, 2005
4. ### yeziGuest

thanks, I will first read the FAQ .

yezi, Nov 29, 2005
5. ### Gordon BurdittGuest

>I have question with the data type problem. According to the C
>standard. IEEE 754
>
>float 4 3.4x10-38E..3.4x10+38E
>double 8 1.7x10-308E..1.7x10+308E
>long double 12 ???
>
>The code function is (0.000200-0.000300)*(0.000200-0.000300)
>suppose the result (difference ) should be the 10 (-8)
>
>difference I claim is double;
>
>But THe code result is :
>fp1 0.000200
>------------------------
>fp2 0.000300
>difference is 0.000000

>which means the result is zero.

You print out a value you expect to be 1.0e-8 with *six* digits
after the decimal point, and you wonder why it shows as zero?

Try printing the result with a format like %200.100f . You'll
notice that the result isn't exact. None of the following numbers
are exactly representable in binary floating point: 0.000200,
0.000300, 1.0e-8. For that matter, *most* aren't unless they are
exact integers. A few exceptions are: 0.5, 0.25, 0.75, 0.125,
0.375, 0.625, 0.875 .

Gordon L. Burditt

Gordon Burditt, Nov 29, 2005
6. ### Keith ThompsonGuest

"yezi" <> writes:
> thanks, I will first read the FAQ .

Please read this as well:

<http://cfaj.freeshell.org/google/>

(Thanks to Chris F.A. Johnson for setting up this page.)

--
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, Nov 29, 2005
7. ### Richard HeathfieldGuest

Keith Thompson said:

> "yezi" <> writes:
>> thanks, I will first read the FAQ .

>
> Please read this as well:
>
> <http://cfaj.freeshell.org/google/>
>
> (Thanks to Chris F.A. Johnson for setting up this page.)
>

Just a nit...

<li><a
href="http://www.catb.org/~esr/faqs/smart-questions.html#bespecific">http://www.catb.org/~esr/faqs/smart-questions.html#bespecifichttp://en.wikipedia.org/wiki/Netiquette</a>

is a bit longer than it should be.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Richard Heathfield, Nov 29, 2005
8. ### jacob naviaGuest

yezi wrote:

> Hi: All:
>
> I have question with the data type problem. According to the C
> standard. IEEE 754
>
> float 4 3.4x10-38E..3.4x10+38E
> double 8 1.7x10-308E..1.7x10+308E
> long double 12 ???
>
> The code function is (0.000200-0.000300)*(0.000200-0.000300)
> suppose the result (difference ) should be the 10 (-8)
>
> difference I claim is double;
>
> But THe code result is :
> fp1 0.000200
> ------------------------
> fp2 0.000300
> difference is 0.000000
>
> which means the result is zero.
>
> Why?
>
> Thanks
> bin YE
>

Using lcc-win32 I type:
#include <stdio.h>
int main(void)
{
double f = 0.0002;
double g = 0.0003;
double delta = (f-g)*(f-g);
printf("%e\n",delta);
}

D:\lcc\mc63\test> lc texact.c
D:\lcc\mc63\test> texact
1.000000e-008

D:\lcc\mc63\test>

You are using the wrong printf format.

jacob navia, Nov 29, 2005
9. ### yeziGuest

This is work with %e, I find the type declarion and printf in sort of
complex . It should be matched very well then got u want .

yezi, Nov 29, 2005
10. ### Default UserGuest

yezi wrote:

> This is work with %e, I find the type declarion and printf in sort of
> complex . It should be matched very well then got u want .

What is?

Brian

--
Please quote enough of the previous message for context. To do so from
Google, click "show options" and use the Reply shown in the expanded
header.

Default User, Nov 29, 2005
11. ### Keith ThompsonGuest

"yezi" <> writes:
> This is work with %e, I find the type declarion and printf in sort of
> complex . It should be matched very well then got u want .

Please read this:

<http://cfaj.freeshell.org/google/>

--
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, Nov 30, 2005

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