Calculating the average?

Question

I have a method that calculates the average

Note class:

private Double moyenneFinale;

@Column(name = "moyenne_finale")
public Double getMoyenneFinale() {
    return this.moyenneFinale;
}

public void setMoyenneFinale(Double moyenneFinale) {
    this.moyenneFinale = moyenneFinale;
}

Class calBean

private double result;
public double moyenneFinale;
public calNote {
    result=0.0
    result= ((100 * 20)/100)/45;
    moyenneFinale=result;
    note.setMoyenneFinale(moyenneFinale);
}
//getter and setter

the result should be 0.44 but recorded in the database given value is 0.0 if I make +1 it will be 1.0

possible duplicate of [Division of integers in Java](http://stackoverflow.com/questions/7220681/division-of-integers-in-java) — BackSlash, May 07 '14 at 10:13
@BackSlash The question may be a duplicate but not exactly of the question you linked since this question adds the nuance of operands being literals and their type being changed by adding a `.0` to them. — Pablo Francisco Pérez Hidalgo, May 07 '14 at 10:30

Pablo Francisco Pérez Hidalgo · Accepted Answer · 2014-05-07T10:20:36.177

2

All the numeric literals at ((100 * 20)/100)/45 are integers so each operation performs integer arithmetic instead floating pointer arithmetic and finally the result is cast to double at the assignation operation.

You should change at least one of your literals to 100.0, 20.0, 45.0, ...

edited May 07 '14 at 10:20

answered May 07 '14 at 10:15

Pablo Francisco Pérez Hidalgo

27,044
8
36
62

_You should change your constants to 100.0, 20.0, 45.0_ Not exactly: Changing just one of them will do the trick – BackSlash May 07 '14 at 10:19
@BackSlash That is true, I'll edit my answer to make it more clear. – Pablo Francisco Pérez Hidalgo May 07 '14 at 10:20
@BackSlash I think that changing all of them is clearer and safer. – cangrejo May 07 '14 at 10:21
@PabloFranciscoPérezHidalgo Those are called literals, not constants. – cangrejo May 07 '14 at 10:21
@broncoAbierto Can you elaborate? Why do you think it is safer? – BackSlash May 07 '14 at 10:22
@BackSlash Because if you were to change the expression and eliminate the one `double` literal, you could forget to change another one of them to keep the desired result. – cangrejo May 07 '14 at 10:24

Rod · Answer 2 · 2014-05-07T10:45:17.347

In line

result= ((100 * 20)/100)/45;

you are using operator / on integers. When used on integers, operator / behaves as "DIV" (resulting integer). Examples: 10/4=2, but 10.0/4=2.5, 10/4.0=2.5, 10.0/4.0=2.5.

Options are:

To set numbers as 100.0, 20.0, 100.0 and 45.0. (It is enough to set just first one to 100.0).

result= ((100.0 * 20)/100)/45;

To use casting on numbers. (double)100.

result= (((double)100 * 20)/100)/45;

Or to add 1.0* before equation;

result= 1.0*((100 * 20)/100)/45;

Tho, first method is best in your case, second and third are useful if you are given integers x, y, z, t and your result is ((x * y)/z)/t;

Calculating the average?

2 Answers2