a very weird result

[ai lian]

The code is as the following:

{
double limit= 0.02*33.00;
double limit1= 0.06*11.00;
double limit2= 0.03*22.00;
double limit3= 0.01*66.00;

limit=(floor(limit*100))/100;
limit1=(floor(limit1*100))/100;
limit2=(floor(limit2*100))/100;
limit3=(floor(limit3*100))/100;

printf("limit= %.2f\n",limit);
printf("limit1 = %.2f\n",limit1);
printf("limit2 = %.2f\n",limit2);
printf("limit3 = %.2f\n",limit3);
}

What's weird is the output:

limit= 0.66
limit1 = 0.65
limit2 = 0.65
limit3 = 0.66

Why are the result for limit1 and limit2 is 0.65 instead of 0.66?

Thanks in advance.

 

[Karl Heinz Buchegger]

The code is as the following:

{
double limit= 0.02*33.00;
double limit1= 0.06*11.00;
double limit2= 0.03*22.00;
double limit3= 0.01*66.00;

limit=(floor(limit*100))/100;
limit1=(floor(limit1*100))/100;
limit2=(floor(limit2*100))/100;
limit3=(floor(limit3*100))/100;

printf("limit= %.2f\n",limit);
printf("limit1 = %.2f\n",limit1);
printf("limit2 = %.2f\n",limit2);
printf("limit3 = %.2f\n",limit3);
}

What's weird is the output:

limit= 0.66
limit1 = 0.65
limit2 = 0.65
limit3 = 0.66

Why are the result for limit1 and limit2 is 0.65 instead of 0.66?

Thanks in advance.

See eg.
http://www.petebecker.com/js200006.html

 

[John Harrison]

The code is as the following:

{
double limit= 0.02*33.00;
double limit1= 0.06*11.00;
double limit2= 0.03*22.00;
double limit3= 0.01*66.00;

limit=(floor(limit*100))/100;
limit1=(floor(limit1*100))/100;
limit2=(floor(limit2*100))/100;
limit3=(floor(limit3*100))/100;

printf("limit= %.2f\n",limit);
printf("limit1 = %.2f\n",limit1);
printf("limit2 = %.2f\n",limit2);
printf("limit3 = %.2f\n",limit3);
}

What's weird is the output:

limit= 0.66
limit1 = 0.65
limit2 = 0.65
limit3 = 0.66

Why are the result for limit1 and limit2 is 0.65 instead of 0.66?

Thanks in advance.

Rounding errors.

0.06*11.00*100 does not equal 66, obviously it equals a number very
slightly less than 66, floor then rounds that number down to 65 and so you
get 0.65.

Never assume that floating point arithmetic will give you exact results,
in particular your problem is that 0.06 is not *exactly* 6 one hundredths,
and the rounding errors start there.

john