21

I can see this is a common problem for new programmers, however I didn't succeed in implementing any of the solution to my code. Basically I want to divide w and v, which must be saved to a double variable. But it prints [0.0, 0.0, ... , 0.0]

public static double density(int[] w, int[] v){
double d = 0;
    for(L = 0; L < w.length; L++){
        d = w[L]  /v[L];
    }
    return d;
}
roarknuppel
  • 269
  • 1
  • 2
  • 8
  • 2
    Google: integer division. It will give you an integer result with truncation. Write d = (double)w[L] /v[L]; – Bathsheba Sep 30 '13 at 09:18
  • Possible duplicate of [Why is the result of 1/3 == 0?](http://stackoverflow.com/questions/4685450/why-is-the-result-of-1-3-0) – fabian Mar 06 '16 at 20:34
  • If precision is needed, do not use primitive double or its wrapper class. Use `BigDecimal` instead. – hfontanez Feb 09 '22 at 20:51

3 Answers3

38

This line here d = w[L] /v[L]; takes place over several steps

d = (int)w[L]  / (int)v[L]
d=(int)(w[L]/v[L])            //the integer result is calculated
d=(double)(int)(w[L]/v[L])    //the integer result is cast to double

In other words the precision is already gone before you cast to double, you need to cast to double first, so

d = ((double)w[L])  / (int)v[L];

This forces java to use double maths the whole way through rather than use integer maths and then cast to double at the end

Richard Tingle
  • 16,906
  • 5
  • 52
  • 77
5

use like following

    d = (double) w[L]  /v[L];
stinepike
  • 54,068
  • 14
  • 92
  • 112
3

Actaully w[L]/v[L] here both are integers. On the division operation it lost precision and truncated to integer value. Then the truncated integer is converted to double.

The solution would be convert the operand or the divisor or both to a double. Then the division operation would produce a double.

d = w[L]/(double)v[L];
will
  • 23
  • 5
Subhrajyoti Majumder
  • 40,646
  • 13
  • 77
  • 103