Double Value error in C

Question

I am having a bit problem with this simple code.

I have tried different type castings, but I have not got any issue in here. If anybody tell me the reason that why its always printing 0.0000000 or something like that, I will be regretful and thanks in advance.

#include<stdio.h>
main()
{
    double num1,num5;
    printf ("Hello user\n");
    printf ("Let the thickness of black and white brick are same and the value 
             is 2m\nAnd the gap between them is 1m\n");
    printf ("\nEnter the length of the road:\n");
    scanf ("%lf",&num1);
    num5 = ((2/5)*num1);
    printf ("%lf",num5);
    return 0;
}

Duplicate with https://stackoverflow.com/questions/40443602/c-double-not-working-as-expect/40443639#40443639 and https://stackoverflow.com/questions/16221776/why-dividing-two-integers-doesnt-get-a-float — Guillaume Jacquenot, Jun 04 '17 at 20:50
Possible duplicate of [Dividing 1/n always returns 0.0](https://stackoverflow.com/questions/13331054/dividing-1-n-always-returns-0-0) — phuclv, Jun 05 '17 at 06:57

Weather Vane · Answer 1 · 2017-06-04T17:32:06.677

2

(2/5) performs an integer calculation, result 0. I suggest one of these approaches

num5 = 0.4 * num1;

num5 = 2 * num1 / 5;

The second is potentially more accurate (since 0.4 cannot be exactly represented in double) when num1 is already a multiple of 5. In the first case, an error is already there.

edited Jun 04 '17 at 17:32

answered Jun 04 '17 at 17:28

Weather Vane

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Or `num5 = (2.0 / 5) * num1;`? – pmg Jun 04 '17 at 17:31
Or `num5 = (2 / 5.0) * num1;` but they might actually be integer variables in a real example. – Weather Vane Jun 04 '17 at 17:33

score 1 · Answer 2 · answered Jun 04 '17 at 17:27

In integer math, 2/5 is 0 (remainder two). In C, how you use a result does not affect how that result is computed (the rules are complex enough already). So the fact that you multiply the result by a double doesn't change the way 2/5 (the quotient of two integers) is computed.

score 0 · Answer 3 · answered Jun 04 '17 at 17:48

0

2 and 5 are integers they are not double. Adding decimal point will yield the correct result.

Something like this.

num5 = ((2.0/5.0)* num1);

or

  num5 = ((2.0/5)* num1);

or

  num5 = ((2/5.0)* num1);

answered Jun 04 '17 at 17:48

danglingpointer

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Double Value error in C

3 Answers3