CODEWITHSUNDEEP
cfloating-pointrounding
couscousman
couscousman

Reputation: 84

Why is the function floor giving different results in this case?

In this example, the behaviour of floor differs and I do not understand why:

printf("floor(34000000.535 * 100 + 0.5) : %lf \n", floor(34000000.535 * 100 + 0.5));
printf("floor(33000000.535 * 100 + 0.5) : %lf \n", floor(33000000.535 * 100 + 0.5));

The output for this code is:

floor(34000000.535 * 100 + 0.5) : 3400000053.000000
floor(33000000.535 * 100 + 0.5) : 3300000054.000000

Why does the first result not equal to 3400000054.0 as we could expect?

Upvotes: 0

Views: 166

Answers (2)

chux
chux

Reputation: 153517

double in C does not represent every possible number that can be expressed in text.

double can typically represent about 264 different numbers. Neither 34000000.535 nor 33000000.535 are in that set when double is encoded as a binary floating point number. Instead the closest representable number is used.

Text             34000000.535
closest double   34000000.534999996423...
Text             33000000.535
closest double   33000000.535000000149...

With double as a binary floating point number, multiplying by a non-power-of-2, like 100.0, can introduce additional rounding differences. Yet in these cases, it still results in products, one just above xxx.5 and another below.

Adding 0.5, a simple power of 2, does not incurring rounding issues as the value is not extreme compared to 3x00000053.5.

Seeing intermediate results to higher print precision well shows the typical step-by-step process.

#include <stdio.h>
#include <float.h>
#include <math.h>

 void fma_test(double a, double b, double c) {
   int n = DBL_DIG + 3;
   printf("a b c      %.*e %.*e %.*e\n", n, a, n, b, n, c);
   printf("a*b        %.*e\n", n, a*b);
   printf("a*b+c      %.*e\n", n, a*b+c);
   printf("a*b+c      %.*e\n", n, floor(a*b+c));
   puts("");
 }

int main(void) {
  fma_test(34000000.535, 100, 0.5);
  fma_test(33000000.535, 100, 0.5);
}

Output

a b c      3.400000053499999642e+07 1.000000000000000000e+02 5.000000000000000000e-01
a*b        3.400000053499999523e+09
a*b+c      3.400000053999999523e+09
a*b+c      3.400000053000000000e+09

a b c      3.300000053500000015e+07 1.000000000000000000e+02 5.000000000000000000e-01
a*b        3.300000053500000000e+09
a*b+c      3.300000054000000000e+09
a*b+c      3.300000054000000000e+09

The issue is more complex then this simple answers as various platforms can 1) use higher precision math like long double or 2) rarely, use a decimal floating point double. So code's results may vary.

Upvotes: 3

Peter Hrvola
Peter Hrvola

Reputation: 9

Question has been already answered here.

In basic float numbers are just approximation. If we have program like this:

float a = 0.2 + 0.3;
float b = 0.25 + 0.25;

if (a == b) {
    //might happen
}
if (a != b) {
    // also might happen
}

The only guaranteed thing is that a-b is relatively small.

Upvotes: 0

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