Why 5.0/3 is 1.666667 in C?

Question

I'm new in C language, but I've tried integer, float and double division in C as I'm normally doing in Java, but when I execute 5.0/3 instead of 1.6666666666666667 I'm getting 1.666667 for double division and for float division.

I had tried to execute the program using Visual Studio as I always do but I got the message "First number is 1, second one is 1.666667 and the last one is 1.666667." after executing:

#include <stdio.h>

int main()
{
    int firstNumber = 5 / 3;
    float secondNumber = 5.0f / 3.0f;
    double thirdNumber = 5.0 / 3.0;

    printf("First number is %d, second one is %f and the last one is %lf.", firstNumber, secondNumber, thirdNumber);
    return 0;
}

Why I'm getting the same result for 'secondNumber' and for 'thirdNumber'?

Possible duplicate of [Is floating point math broken?](https://stackoverflow.com/questions/588004/is-floating-point-math-broken) — Marco Bonelli, Jul 06 '19 at 00:08
Ppl, this question has nothing to do with marked "dupe", the question is legit and seems original. — qrdl, Jul 06 '19 at 05:41

chux - Reinstate Monica · Accepted Answer · 2019-07-06T00:55:24.113

Typical float can represent about 2³² different values.

Typical double can represent about 2⁶⁴ different values.

In both types, ⁵/₃, the exact quotient of the division, is not in that set. Instead a nearby value (some binary fraction) is used.

float secondNumber = 5.0f / 3.0f; // 1.66666662693023681640625
double thirdNumber = 5.0 / 3.0;   // 1.6666666666666667406815349750104360282421112060546875

When using "%f", 6 places past the decimal point are used. The printed text is a rounded one. In both cases, rounding to the same.

1.666667

To see more digits, use "%.10f", "%.20f", etc. @xing

printf("%.10f\n", secondNumber);
printf("%.10f\n", thirdNumber);

Output

1.6666666269
1.6666666667

Why 5.0/3 is 1.666667 in C?

1 Answers1