Loss of precision on adding doubles?

Question

folks! I've encountered a little problem: I'm doing a simple addition with three double values. The result has a smaller precision than the used values.

double minutes = 3;
minutes = minutes / (24.0*60.0);  // contains 0.00208333
double hours = 3;
hours = hours / 24.0; // contains 0.125
double days = 3; // contains 3 

double age = days + hours + minutes; // result is 3.12708

I found no way to avoid this behaviour.

What is the problem? Calculation seems ok. How do you print result? Do you want more digits printed? — zch, Jun 20 '13 at 12:21
note that: 1) depending on the way you output your values, you may not see all significant digits and 2) doubles are not able to exactly represent values like 0.1 — PlasmaHH, Jun 20 '13 at 12:22
Re: precision: all four print six significant digits. Where is the difference? — Thilo, Jun 20 '13 at 12:22
possible duplicate-- http://stackoverflow.com/questions/554063/how-do-i-print-a-double-value-with-full-precision-using-cout — 0decimal0, Jun 20 '13 at 13:13

score 5 · Accepted Answer · answered Jun 20 '13 at 12:26

5

Nothing seems to be wrong with the calculation as what the comments on your post said.

If you'd like to see more precision consider looking up setprecision()

answered Jun 20 '13 at 12:26

Chris Wijaya

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The calculation is correct, indeed. I validated this by using printf(). The loss of precision was been caused by outputting it by std::cerr or assigning it to a std::stringstream. setprecision did the trick. Thanks a lot! – XQDev Jun 20 '13 at 12:58

score 0 · Answer 2 · answered Jun 20 '13 at 12:52

0

There is no problem. The significant figure of 0.00208333 and 3.12708 are both 6. This is a correct result.

answered Jun 20 '13 at 12:52

qpalz

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In my case it WAS a problem, even if this may be no misbehaviour. ;) – XQDev Jun 20 '13 at 13:01

0decimal0 · Answer 3 · 2013-08-11T04:10:15.923

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Sometimes in order to get the value according to the desired precision we have to specify the limit of precision,here is my code which works fine ,hope it helps:

double minutes = 3;
minutes = minutes / (24.0*60.0);
double hours = 3;
hours = hours / 24.0; 
double days = 3;
double age = days + hours + minutes;
printf("%.8f",age);//here i have included the places of precision (.8)

edited Aug 11 '13 at 04:10

answered Jun 20 '13 at 12:59

0decimal0

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Loss of precision on adding doubles?

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