Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
24  3257 
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
On Jul 23, 5:28 pm, David Mark <dmark.cins...@ gmail.comwrote:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
very good. thank you.
David Mark said the following on 7/23/2007 5:28 PM:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
>Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
0.00831 + 0.000001 != 8311
--
Randy
Chance Favors The Prepared Mind
comp.lang.javas cript FAQ - http://jibbering.com/faq/index.html
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/
On Jul 23, 6:08 pm, Randy Webb <HikksNotAtH... @aol.comwrote:
David Mark said the following on 7/23/2007 5:28 PM:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
0.00831 + 0.000001 != 8311
To compare the result, as per my post, you have to multiply both sides
of the equation. I don't know how much more explicit I could have
made this and the OP obviously understood.
David Mark said:
>
On Jul 23, 6:08 pm, Randy Webb <HikksNotAtH... @aol.comwrote:
>David Mark said the following on 7/23/2007 5:28 PM:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
0.00831 + 0.000001 != 8311
To compare the result, as per my post, you have to multiply both sides
of the equation. I don't know how much more explicit I could have
made this and the OP obviously understood.
Saying "multiply each ... before adding" suggests that the things
referred to by "each" are also the things that you are adding, so
you certainly could have been much more clear about specifying
that you also have to multiply the comparison target.
How is it obvious that the OP understood? For all we know, he's
currently scratching his head because your solution, which seems
so simple, doesn't work when implemented.
--
How is it obvious that the OP understood?
Let's see: 0.00831 + 0.000001 = 0.008311
(0.00831*100000 0) + (0.000001*10000 00) == 0.008311/1000000
8310 + 1 = 8311/1000000
thus ==0.008311
What does everyone think? I'd appreciate a better way of handling this
if there is one...
Thanks
On Jul 23, 6:31 pm, Lee <REM0VElbspamt. ..@cox.netwrote :
David Mark said:
On Jul 23, 6:08 pm, Randy Webb <HikksNotAtH... @aol.comwrote:
David Mark said the following on 7/23/2007 5:28 PM:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
0.00831 + 0.000001 != 8311
To compare the result, as per my post, you have to multiply both sides
of the equation. I don't know how much more explicit I could have
made this and the OP obviously understood.
Saying "multiply each ... before adding" suggests that the things
referred to by "each" are also the things that you are adding, so
Of course they are.
you certainly could have been much more clear about specifying
that you also have to multiply the comparison target.
How much more clear could the example be? It explicitly shows the new
equation.
>
How is it obvious that the OP understood? For all we know, he's
Because he said so. And now two other posters have muddled the issue
and confused him. How do I know that? Read the thread.
currently scratching his head because your solution, which seems
No. Read the thread.
so simple, doesn't work when implemented.
The example equation "works" fine, in that both sides are equal.
David Mark said the following on 7/23/2007 7:24 PM:
On Jul 23, 6:31 pm, Lee <REM0VElbspamt. ..@cox.netwrote :
>David Mark said:
>>On Jul 23, 6:08 pm, Randy Webb <HikksNotAtH... @aol.comwrote:
David Mark said the following on 7/23/2007 5:28 PM:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
>Is there a good solution for floating-point errors?
>If I got:
>0.00831 + 0.000001 = 0.0083110999999 9999999
>can I avoid this by a better method than comparing 2 number length
>after decimal point and using toFixed() to match shorter number with
>another one?
>Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
0.00831 + 0.000001 != 8311
To compare the result, as per my post, you have to multiply both sides
of the equation. I don't know how much more explicit I could have
made this and the OP obviously understood.
Saying "multiply each ... before adding" suggests that the things
referred to by "each" are also the things that you are adding, so
Of course they are.
>you certainly could have been much more clear about specifying
that you also have to multiply the comparison target.
How much more clear could the example be? It explicitly shows the new
equation.
Answer:
Multiply both of your decimals by 100000 (in this case), add them, then
divide the answer by 100000 to get the decimal back.
You wanted more clear, you got it.
--
Randy
Chance Favors The Prepared Mind
comp.lang.javas cript FAQ - http://jibbering.com/faq/index.html
Javascript Best Practices - http://www.JavascriptToolbox.com/bestpractices/
On Jul 23, 7:35 pm, Randy Webb <HikksNotAtH... @aol.comwrote:
David Mark said the following on 7/23/2007 7:24 PM:
On Jul 23, 6:31 pm, Lee <REM0VElbspamt. ..@cox.netwrote :
David Mark said:
>On Jul 23, 6:08 pm, Randy Webb <HikksNotAtH... @aol.comwrote:
David Mark said the following on 7/23/2007 5:28 PM:
On Jul 23, 4:35 pm, vunet...@gmail. com wrote:
Is there a good solution for floating-point errors?
If I got:
0.00831 + 0.000001 = 0.0083110999999 9999999
can I avoid this by a better method than comparing 2 number length
after decimal point and using toFixed() to match shorter number with
another one?
Thanks.
Yes. To make it simpler, multiply each by 1000000 before adding:
8310 + 1 = 8311
0.00831 + 0.000001 != 8311
To compare the result, as per my post, you have to multiply both sides
of the equation. I don't know how much more explicit I could have
made this and the OP obviously understood.
Saying "multiply each ... before adding" suggests that the things
referred to by "each" are also the things that you are adding, so
Of course they are.
you certainly could have been much more clear about specifying
that you also have to multiply the comparison target.
How much more clear could the example be? It explicitly shows the new
equation.
Answer:
Multiply both of your decimals by 100000 (in this case), add them, then
divide the answer by 100000 to get the decimal back.
You've done it again. There are six decimal places to shift (in this
case), not five. 100000 != 1000000.
And what the hell does "get the decimal back" mean anyway? Same for
"multiply both of your decimals."
You wanted more clear, you got it.
I didn't want anything. The OP wanted an answer and got it and
followed up to say it worked. Now you have made a real mess of
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