Float arithmetic operation yields value = 0

Question

The below program yields value = 0.000000, although the format specifier is that of a float.

#include <stdio.h>
int main ()
{
        printf ("%f", 5/9);
        return 0;
}

`When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded` — Jayesh Bhoi, Sep 19 '14 at 10:10

score 3 · Answer 1 · answered Sep 19 '14 at 10:07

The expression 5/9 has two integer arguments and so is evaluated used integer division. Hence the result is 0.

You then invoke undefined behaviour by passing an int to a %f format specifier.

Change at least one of the operands to a floating point value to use floating point division:

printf("%f", 5.0/9.0);

score 1 · Answer 2 · answered Sep 19 '14 at 10:08

1

 printf ("%f", 5/9);

5 / 9 is an integer division. The expression yields an int but f requires a float or a double argument.

Change the call to:

printf ("%f", 5 / 9.0);

to perform a floating-point division and have a double argument.

answered Sep 19 '14 at 10:08

ouah

score 0 · Answer 3 · answered Sep 19 '14 at 10:11

0

5 and 9 is integer. Division of two integer will be always integer. Make one of them a float or a double

 printf ("%f", 5.0/9);

or

 printf ("%f", 5/9.0);

answered Sep 19 '14 at 10:11

Deb

3 Answers3