Division result is always zero [duplicate]

Question

I got this C code.

#include <stdio.h>

int main(void)
{
        int n, d, i;
        double t=0, k;
        scanf("%d %d", &n, &d);
        t = (1/100) * d;
        k = n / 3;
        printf("%.2lf\t%.2lf\n", t, k);
        return 0;
}

I want to know why my variable 't' is always zero (in the printf function) ?

Bedding

See also #1580832 – Ephemerality

lol - integer division in c was my very first head scratcher too. No stack over flow then - just a humorless prof – Monthly

Answer 1

because in this expression

t = (1/100) * d;

1 and 100 are integer values, integer division truncates, so this It's the same as this

t = (0) * d;

you need make that a float constant like this

t = (1.0/100.0) * d;

you may also want to do the same with this

k = n / 3.0;

Farcy

Or just use d / 100.0. – Sundae

In my situation this is what I ended up doing to stop getting 0 for my division: Double ratio = (Convert.ToDouble(x) / Convert.ToDouble(y)); – Pythagoreanism

@ChristopherD.Emerson That is C#; this question is about C (and also applies to C++). – Collette

Answer 2

You are using integer division, and 1/100 is always going to round down to zero in integer division.

If you wanted to do floating point division and simply truncate the result, you can ensure that you are using floating pointer literals instead, and d will be implicitly converted for you:

t = (int)((1.0 / 100.0) * d);

Chaffee

Answer 3

I think its because of

t = (1/100) * d;

1/100 as integer division = 0

then 0 * d always equals 0

if you do 1.0/100.0 i think it will work correctly

Massa

Answer 4

t = (1/100) * d; 

That is always equals 0,you can do this

t=(1%100)*d 

and add it to 0

Jecoa