multiplication weirdness

[harborcoat-design]

Hello -

I thought I was reasonably versed in JavaScript, buuuuut...

When I do the following:

x = parseFloat(326.655);
y = parseFloat(100);
z = x * y;
alert(z);

When I should get 32665.5 I instead get 32665.499999999996.

Why is that?

Thanks for any help.

Tom

[web.dev]

harborcoat-design wrote:

Hello -

I thought I was reasonably versed in JavaScript, buuuuut...

When I do the following:

x = parseFloat(326.655);
y = parseFloat(100);
z = x * y;
alert(z);

When I should get 32665.5 I instead get 32665.499999999996.

Why is that?

Thanks for any help.

Tom

Past postings might help:

http://groups.google.com/group/comp....5f3e157c202fd1

http://groups.google.com/group/comp....d36142924ccc21

[Evertjan.]

harborcoat-design wrote on 09 nov 2005 in comp.lang.javascript:

I thought I was reasonably versed in JavaScript, buuuuut...
Seems you are not, sorry.
When I do the following:

x = parseFloat(326.655);
y = parseFloat(100);
z = x * y;
alert(z);
The same goes for:

x = 326.655;
y = 100;
z = x * y;
alert(z);

So why the parseFloat()?
When I should get 32665.5 I instead get 32665.499999999996.
No, you should not.
Why is that?

Javascript stores floating point values in a binary format,
and this is what is the nearest result of your multiplication
in binary floating point outputted in a decimal format.

--
Evertjan.
The Netherlands.
(Replace all crosses with dots in my emailaddress)

[Michael Winter]

On 09/11/2005 18:01, harborcoat-design wrote:

I thought I was reasonably versed in JavaScript, buuuuut...
[snip]
x = parseFloat(326.655);
y = parseFloat(100);
Using the parseFloat function here is indeed something to question. The
parseFloat function operates on string values, and should only be
necessary for stripping off non-digit characters; use unary plus (+)
when you know the value is a properly formatted number.

[snip]
When I should get 32665.5 I instead get 32665.499999999996.

Reading the FAQ really can save you, and others, time.

Mike
c.l.javascript FAQ: <http://www.jibbering.com/faq/>

--
Michael Winter
Prefix subject with [News] before replying by e-mail.

[harborcoat-design]

Seems you are not, sorry.
That's okay - I'm man enough to take my medicine.
Javascript stores floating point values in a binary format,
and this is what is the nearest result of your multiplication
in binary floating point outputted in a decimal format.
Yes - I see - Thanks for the reply.
--
Evertjan.
The Netherlands.
(Replace all crosses with dots in my emailaddress)

[Evertjan.]

Michael Winter wrote on 09 nov 2005 in comp.lang.javascript:

use unary plus (+)
when you know the value is a properly formatted number.

Not neccessary in this case with a following multiplication.

--
Evertjan.
The Netherlands.
(Replace all crosses with dots in my emailaddress)

[Michael Winter]

On 09/11/2005 18:24, Evertjan. wrote:

Michael Winter wrote on 09 nov 2005 in comp.lang.javascript:

use unary plus (+) when you know the value is a properly formatted
number.

Not neccessary in this case with a following multiplication.

I was referring to type-conversion in general, not the OP's specific
case. No conversion was necessary at all, there.

Mike

--
Michael Winter
Prefix subject with [News] before replying by e-mail.

[Dr John Stockton]

JRS: In article <11*********************@z14g2000cwz.googlegroups. com>,
dated Wed, 9 Nov 2005 10:01:45, seen in news:comp.lang.javascript,
harborcoat-design <tj******@gmail.com> posted :

I thought I was reasonably versed in JavaScript, buuuuut...
Without reading the newsgroup FAQ, as posted every Mon & Fri ??
When I do the following:

x = parseFloat(326.655);
y = parseFloat(100);
z = x * y;
alert(z);

When I should get 32665.5 I instead get 32665.499999999996.

Why is that?

After the FAQ, read <URL:http://www.merlyn.demon.co.uk/js-maths.htm> and
links.

Note that 100 will be stored exactly, but it is not a power of 2.
326.655 cannot be stored exactly, so multiplication by 100 loses some
bits of the ideal result, giving something that outputs as you see.

But 0.01 cannot be stored exactly, so dividing by 0.01 will not
*necessarily* give the same result, although in this case it does.

AFAICS, parseFloat is worth using only when its parameter can be a
string with other characters after the number.

--
© John Stockton, Surrey, UK. ?@merlyn.demon.co.uk Turnpike v4.00 IE 4 ©
<URL:http://www.jibbering.com/faq/> JL/RC: FAQ of news:comp.lang.javascript
<URL:http://www.merlyn.demon.co.uk/js-index.htm> jscr maths, dates, sources.
<URL:http://www.merlyn.demon.co.uk/> TP/BP/Delphi/jscr/&c, FAQ items, links.