Re: rounding problems
| От | Justin |
|---|---|
| Тема | Re: rounding problems |
| Дата | |
| Msg-id | 482B09A8.4090806@emproshunts.com обсуждение исходный текст |
| Ответ на | Re: rounding problems (Sam Mason <sam@samason.me.uk>) |
| Ответы |
Re: rounding problems
|
| Список | pgsql-general |
I have forgotten how much i hate C++
Its not doing what you say it would but it did do other odd ball things. I miss my foxpro :-(.
Plus its not holding 15 precision points
#include <stdio.h>
#include <cmath>
int main()
{
double a = 0.1;
//double b = 1000;
double c = 100000000;
double d ; //= a * b;
for( int i = 1 ; i < 10 ; i++)
{
d = pow(a,i)+ c ;
printf("%.10f\n", d);
d = d-c ;
printf("%.10f\n", d);
}
return 0;
}
Sam Mason wrote:
Its not doing what you say it would but it did do other odd ball things. I miss my foxpro :-(.
Plus its not holding 15 precision points
#include <stdio.h>
#include <cmath>
int main()
{
double a = 0.1;
//double b = 1000;
double c = 100000000;
double d ; //= a * b;
for( int i = 1 ; i < 10 ; i++)
{
d = pow(a,i)+ c ;
printf("%.10f\n", d);
d = d-c ;
printf("%.10f\n", d);
}
return 0;
}
Sam Mason wrote:
On Tue, May 13, 2008 at 02:36:18PM -0400, Justin wrote:Double holds 15 places which is the highest value of precision it can maintain before rounding occurs. Is is limit less no, but what is? Practically speaking taking a vale 0.000,000,000,000,001 aka 1 trillionth of anything,But remember that if you add this value onto a large number and then take off the large number the result will be zero. (0.000,000,000,01 + 1,000,000) - 1,000,000 ==> 0 0.000,000,000,01 + (1,000,000 - 1,000,000) ==> 0.000,000,000,01 In general, operations on floating point numbers will increase their errors.i view the problem solved for 98% of problems.Floating point math is good for most problems, hence why most languages expose the abstraction. Sam
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