Lii - 1 year ago 83
C Question

# C: imprecision in arithmetic of double

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 d-(1-e) equal to 0?

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);
}
``````
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