math problem

[Chiwa]

Hey,

Expression:
Math.floor(x * 100) / 100

x= 4.1 gives 4.09, why in gods name?
While other values for x don't give a problem.

Thx in advance

Chiwa

[DJ WIce]

: Expression:
: Math.floor(x * 100) / 100
:
: x= 4.1 gives 4.09, why in gods name?

Because browsers are not calculators (here = sign is "can be").

4.1 = 4.100000000000
= 4.099999999999
= 4.100000000001

So:
4.1*100 = 410.0000000000
= 409.9999999999

So:
Math.floor(x * 100) = 410
= 409

And so your answer can be 409.
How more devisions, multiplications and roundings the bigger the margin to
the exact calculation may grow.

: While other values for x don't give a problem.
No, you just did not find them ;-)

Wouter

[DJ WIce]

: Expression:
: Math.floor(x * 100) / 100

Try this:

x.setPlaces(2);

Wouter

[DJ WIce]

: Try this:
:
: x.setPlaces(2);
See : http://www.mredkj.com/javascript/num...xample150.html

Wouter

[Randy Webb]

DJ WIce wrote:

: Try this:
:
: x.setPlaces(2);
See : http://www.mredkj.com/javascript/num...xample150.html

http://www.jibbering.com/faq/#FAQ4_7

Is a better reference for the OP

--
Randy
Chance Favors The Prepared Mind
comp.lang.javas cript FAQ - http://jibbering.com/faq/

[Brian Genisio]

Chiwa wrote:

Hey,

Expression:
Math.floor(x * 100) / 100

x= 4.1 gives 4.09, why in gods name?
While other values for x don't give a problem.

Thx in advance

Chiwa

Just so you know, this is not just a limitation of Javascript. Many
other languages (C/C++ for instance) have this as an issue. That is
because Real numbers (floating point numbers) are _ALWAYS_
approximations. When you do a measurement of your finger (for
instance), and you measure exactly 1.25 inches... is that really
correct? Or could it be 1.2500000234358 9 inches instead? Or
1.2500000034322 349872342229? (You get the idea)

For this reason (and others dealing with the complexity of floating
point numbers), when a float (indescrete) is represented by binary data
(descrete), you will have some goofiness... it is guaranteed.

Because of this, most languages have methods for you to set the
precision (the number of significant figures), where rounding will occur
to the precision value you wish for.

If you desire an exact floating point value, such as in something like
money calculations, simply set your precisions.

You can also work in values multiplied by (10*order)... For instance,
$10.47 can be worked with as integers... immagine 1,047 pennies :) You
can convert it back by dividing or modding by (10*order) in integer form.

I have had to use that method once when developing assembly code on an
8-bit processor. There was no float support, so I needed to work in
values multiplied by (10*order), and then do some simple calculations to
provide useful user information.

Blah.... I wrote too much. Oh well, I hope it helps :)
Brian

[Lasse Reichstein Nielsen]

"Chiwa" <Ch******@hotma il.com> writes:

Expression:
Math.floor(x * 100) / 100

x= 4.1 gives 4.09, why in gods name?
While other values for x don't give a problem.

<URL:http://jibbering.com/faq/#FAQ4_7>

/L
--
Lasse Reichstein Nielsen - lr*@hotpop.com
DHTML Death Colors: <URL:http://www.infimum.dk/HTML/rasterTriangleD OM.html>
'Faith without judgement merely degrades the spirit divine.'

[Cardman]

On Sun, 01 Feb 2004 21:25:58 GMT, "Chiwa" <Ch******@hotma il.com>
wrote:

Hey,

Expression:
Math.floor(x * 100) / 100

x= 4.1 gives 4.09, why in gods name?
While other values for x don't give a problem.

Thx in advance

After working on my order form code it appears that I noticed and
solved this same problem long ago.

What I do is to add 0.000001 to the value x before putting it through
this same function, when this makes sure that the value is not below
the true value before you run the calculation.

Since this fraction is so small, then it would be rounded off
anywhere, where you only have to watch out that your x value does not
use this many decimal places.

That of course is not a problem for money values, like what I use it
for, with only two (or three with VAT) decimal places.

There you go, perfect rounding results every time.

Cardman
http://www.cardman.com
http://www.cardman.co.uk

[Brian Genisio]

Cardman wrote:

After working on my order form code it appears that I noticed and
solved this same problem long ago.

What I do is to add 0.000001 to the value x before putting it through
this same function, when this makes sure that the value is not below
the true value before you run the calculation.

Since this fraction is so small, then it would be rounded off
anywhere, where you only have to watch out that your x value does not
use this many decimal places.

That of course is not a problem for money values, like what I use it
for, with only two (or three with VAT) decimal places.

There you go, perfect rounding results every time.

Cardman
http://www.cardman.com
http://www.cardman.co.uk

You have got to be careful with a solution like this, since it is just a
hack, and can be confusing to maintainers in the future. You should
really use precision mechanisms built into the language to solve the
problem more elegantly.

Brian

[Jim Ley]

On Mon, 02 Feb 2004 19:48:00 +0000, Cardman <do****@spam-me.com>
wrote:

That of course is not a problem for money values, like what I use it
for, with only two (or three with VAT) decimal places.

There you go, perfect rounding results every time.

however it is not a legal solution for calculating VAT involving Euros
say. Do it properly.

Jim.
--
comp.lang.javas cript FAQ - http://jibbering.com/faq/