Lii - 5 months ago 25

C Question

I have the following C code:

`int main()`

{

double a=.1,b=.2,c=.3,d=.4;

double e=a+b+c;

if (d- (1-e)){printf("not equal\n");}

printf("%.20f\n",d-(1-e));

}

And the result I get is:

`not equal`

0.00000000000000011102

I know this is due to the imprecision induced from the way computer saves a double. Is there a way to solve this, and make

Answer

As PRP correctly suggests: you need to set a small number to use as zero. The standard C-library (Annex F in the C-standard) offers some macros in `float.h`

for that purpose. You can use them like e.g.:

```
#include <stdio.h>
#include <stdlib.h>
#include <float.h>
#include <math.h>
int main()
{
double a = .1, b = .2, c = .3, d = .4;
double e = a + b + c;
if (d - (1 - e)) {
printf("not equal\n");
}
printf("%.20f\n", d - (1 - e));
printf("%.20f\n", DBL_EPSILON);
if (fabs(d - (1 - e)) <= DBL_EPSILON) {
printf("equal\n");
}
exit(EXIT_SUCCESS);
}
```