exercise Floating point error

Question

I have a list of floating point errors and am having some difficulty with one of the questions. If anyone can help me, I am grateful.

For the code below, I need to justify the reason for the error.

int main () {
  double x;
  x = 0.;
  printf("x = %2.13lf\n",x);
  while ( x != 1.0 ) {
    x = x + 0.1;
    printf("x = %2.13lf\n",x);
 }

I think it is related to the rounding of the sum. Is that right or has something else?

Answer 1

Testing the following code and only by 2.13 for 2.20

#include  <stdio.h>

int main () {
    int i;
    double x;
    i = 1;
    x = 0.;
    printf("exec %1li",i);
    printf("x = %2.20lf\n",x);
    while ( x != 1.0 ) {
        i++;
        x = x + 0.1;
        printf("exec %1li",i);
        printf("x = %2.20lf\n",x);
    }
}

---

The exit to 2.13:

exec 1x = 0.0000000000000
exec 2x = 0.1000000000000
exec 3x = 0.2000000000000
exec 4x = 0.3000000000000
exec 5x = 0.4000000000000
exec 6x = 0.5000000000000
exec 7x = 0.6000000000000
exec 8x = 0.7000000000000
exec 9x = 0.8000000000000
exec 10x = 0.9000000000000
exec 11x = 1.0000000000000 // <- Valor a ser considerado, que na notação 10^13 acaba sendo interpretado como 1.0
exec 12x = 1.1000000000000
exec 13x = 1.2000000000000
exec 14x = 1.3000000000000
exec 15x = 1.4000000000000
exec 16x = 1.5000000000000

---

The exit to 2.20:

exec 1x = 0.00000000000000000000
exec 2x = 0.10000000000000000555
exec 3x = 0.20000000000000001110
exec 4x = 0.30000000000000004441
exec 5x = 0.40000000000000002220
exec 6x = 0.50000000000000000000
exec 7x = 0.59999999999999997780
exec 8x = 0.69999999999999995559
exec 9x = 0.79999999999999993339
exec 10x = 0.89999999999999991118
exec 11x = 0.99999999999999988898 // <- Valor a ser considerado, que na notação 10^13 acaba sendo interpretado como 1.0
exec 12x = 1.09999999999999986677
exec 13x = 1.19999999999999995559
exec 14x = 1.30000000000000004441
exec 15x = 1.40000000000000013323
exec 16x = 1.50000000000000022204

See that at each sum of 0.1 is actually added 0.10000000000000000555, where depending on the number of decimal places used to represent the value, the compiler may end up rounding.

Try changing the line of while for while ( x != 0.5 ) { and you will see that the code will not go into an infinite loop because the value that was added on that last run will literally be 0.5 (0.50000000000000000000 = 0.5 = 0.5000).

I wouldn’t say it’s a rounding error, but rather due to 0.1 represent 0.10000000000000000555 to each sum within the while.