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Can you have a half of any array? [duplicate]

I am an experienced php developer just starting to learn Java. I am following some Lynda courses at the moment and I'm still really early stages. I'm writing sample programs that ask for user input and do simple calculation and stuff.
Yesterday I came across this situation:
double result = 1 / 2;
With my caveman brain I would think result == 0.5, but no, not in Java. Apparantly 1 / 2 == 0.0. Yes, I know that if I change one of the operands to a double the result would also be a double.
This scares me actually. I can't help but think that this is very broken. It is very naive to think that an integer division results in an integer. I think it is even rarely the case.
But, as Java is very widely used and searching for 'why is java's division broken?' doesn't yield any results, I am probably wrong.
My questions are:
Why does division behave like this?
Where else can I expect to find such magic/voodoo/unexpected behaviour?

Java is a strongly typed language so you should be aware of the types of the values in expressions. If not...
1 is an int (as 2), so 1/2 is the integer division of 1 and 2, so the result is 0 as an int. Then the result is converted to a corresponding double value, so 0.0.
Integer division is different than float division, as in math (natural numbers division is different than real numbers division).

You are thinking like a PHP developer; PHP is dynamically typed language. This means that types are deduced at run-time, so a fraction cannot logically produce a whole number, thus a double (or float) is implied from the division operation.
Java, C, C++, C# and many other languages are strongly typed languages, so when an integer is divided by an integer you get an integer back, 100/50 gives me back 2, just like 100/45 gives me 2, because 100/45 is actually 2.2222..., truncate the decimal to get a whole number (integer division) and you get 2.
In a strongly typed language, if you want a result to be what you expect, you need to be explicit (or implicit), which is why having one of your parameters in your division operation be a double or float will result in floating point division (which gives back fractions).
So in Java, you could do one of the following to get a fractional number:
double result = 1.0 / 2;
double result = 1f / 2;
double result = (float)1 / 2;
Going from a loosely typed, dynamic language to a strongly typed, static language can be jarring, but there's no need to be scared. Just understand that you have to take extra care with validation beyond input, you also have to validate types.
Going from PHP to Java, you should know you can not do something like this:
$result = "2.0";
$result = "1.0" / $result;
echo $result * 3;
In PHP, this would produce the output 1.5 (since (1/2)*3 == 1.5), but in Java,
String result = "2.0";
result = "1.0" / result;
System.out.println(result * 1.5);
This will result in an error because you cannot divide a string (it's not a number).
Hope that can help.

I'm by no means a professional on this, but I think it's because of how the operators are defined to do integer arithmetic. Java uses integer division in order to compute the result because it sees that both are integers. It takes as inputs to this "division" method two ints, and the division operator is overloaded, and performs this integer division. If this were not the case, then Java would have to perform a cast in the overloaded method to a double each time, which is in essence useless if you can perform the cast prior anyways.

If you try it with c++, you will see the result is the same.
The reason is that before assigning the value to the variable, you should calculate it. The numbers you typed (1 and 2) are integers, so their memory allocation should be as integers. Then, the division should done according to integers. After that it will cast it to double, which gives 0.0.

Why does division behave like this?
Because the language specification defines it that way.
Where else can I expect to find such magic/voodoo/unexpected behaviour?
Since you're basically calling "magic/voodoo" something which is perfectly defined in the language specification, the answer is "everywhere".
So the question is actually why there was this design decision in Java. From my point of view, int division resulting in int is a perfectly sound design decision for a strongly typed language. Pure int arithmetic is used very often, so would an int division result in float or double, you'd need a lot of rounding which would not be good.

package demo;
public class ChocolatesPurchased
{
public static void main(String args[])
{
float p = 3;
float cost = 2.5f;
p *= cost;
System.out.println(p);
}
}

Java: Is (int) double reliable?

When I perform simple math in java with doubles and other number data types, the double values seem to randomly vary a bit from the supposed result, which might be 5,59999999997 or 6,0000000002 or something. When I cast to int, the double value is obviously rounded down to the next whole number. Does this mean the double could be both 5 or 6? Or does that "5,999999999997" still count as 6 though which would be depending on the binary float value? If not, is there a way to let the negative vary be rounded up, but not lower values from 5,5 to 5,999999999996?
I mean, I dont really want to round the value as described in my last sentence. I'd like to always round down to the next whole number, but I don't want to cause an extra decrement due to wrong double math results.

Converting a double to an int always rounds down. You can round to the nearest whole integer via Math.round(double). The double is varying from what you expect because of floating point error.

If you want to round, you can use the round() method.
double d = 6 +/- some small error
long l = Math.round(d);
Or you can add 0.5 for positive numbers
long l = (long) (d + 0.5);
or
long l = (long) (d + (d < 0 ? -0.5 : 0.5));

I'm not sure I understand the question. Usually when you cast a double to int you add 0.5 to have a nice round.

From the Java Language Specification:
The Java programming language uses round toward zero when converting a floating value to an
integer (§5.1.3), which acts, in this case, as though the number were truncated, discarding
the mantissa bits. Rounding toward zero chooses at its result the format's value closest to
and no greater in magnitude than the infinitely precise result.
So 5,999999999997 when casted to an int will 5 and 6,0000000002 will be 6. If I understand what you are asking with having negative versions of the values (e.g. -5.97), I fail to see how
Math.round() does not suffice you. -6,0000000002 will be rounded to -6 as will -5,999999999997 and every other value above (but not including) -5.5.

Loss of precision after subtracting double from double [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Retain precision with Doubles in java
Alright so I've got the following chunk of code:
int rotation = e.getWheelRotation();
if(rotation < 0)
zoom(zoom + rotation * -.05);
else if(zoom - .05 > 0)
zoom(zoom - rotation * .05);
System.out.println(zoom);
Now, the zoom variable is of type double, initially set to 1. So, I would expect the results to be like 1 - .05 = .95; .95 - .05 = .9; .9 - .05 = .85; etc. This appears to be not the case though when I print the result as you can see below:
0.95
0.8999999999999999
0.8499999999999999
0.7999999999999998
0.7499999999999998
0.6999999999999997
Hopefully someone is able to clearly explain. I searched the internet and I read it has something to do with some limitations when we're storing floats in binary but I still don't quite understand. A solution to my problem is not shockingly important but I would like to understand this kind of behavior.

Java uses IEEE-754 floating point numbers. They're not perfectly precise. The famous example is:
System.out.println(0.1d + 0.2d);
...which outputs 0.30000000000000004.
What you're seeing is just a symptom of that imprecision. You can improve the precision by using double rather than float.
If you're dealing with financial calculations, you might prefer BigDecimal to float or double.

float and double have limited precision because its fractional part is represented as a series of powers of 2 e.g. 1/2 + 1/4 + 1/8 ... If you have an number like 1/10 it has to be approximated.
For this reason, whenever you deal with floating point you must use reasonable rounding or you can see small errors.
e.g.
System.out.printf("%.2f%n", zoom);
To minimise round errors, you could count the number of rotations instead and divide this int value by 20.0. You won't see a rounding error this way, and it will be faster, with less magic numbers.

float and double have precision issues. I would recommend you take a look at the BigDecimal Class. That should take care of precision issues.

Since decimal numbers (and integer numbers as well) can have an infinite number of possible values, they are impossible to map precisely to bits using a standard format. Computers circumvent this problem by limiting the range the numbers can assume.
For example, an int in java can represent nothing larger then Integer.MAX_VALUE or 2^31 - 1.
For decimal numbers, there is also a problem with the numbers after the comma, which also might be infinite. This is solved by not allowing all decimal values, but limiting to a (smartly chosen) number of possibilities, based on powers of 2. This happens automatically but is often nothing to worry about, you can interpret your result of 0.899999 as 0.9. In case you do need explicit precision, you will have to resort to other data types, which might have other limitations.

Using Java doubles (or anything else) to store simple fractions [duplicate]

This question already has answers here:
Division of integers in Java [duplicate]
(7 answers)
Closed 7 years ago.
This seems like a very simple error:
double quarter = 1/4;
Is giving
0.0
Anybody know why this might be happening?
I am trying to store pretty much all the fractions from 1/2 to 1/20 (Just the ones with 1 on the top and in int on the bottom), so I won't be able to input the decimal straight away for all of them.
I've read and heard that floating-point datatypes are not a good way of storing fractions, so is there any other way (in Java)?

Try:
double quarter = 1d/4d;
The division of two integers gives a truncated integer. By putting the d behind the numbers you are casting them to doubles.

For starters, you're trying to divide 1/4 as integer values, and it's truncating it. 1. / 4 will correctly give you 0.25; other ways to express the number 1 as a double include 1d, 1.0, and so on.
Other approaches include:
Use BigDecimal to store values to an exact decimal precision. (For example, this is the preferred way to deal with monetary values.)
Use a Fraction or Rational class, either rolling your own or using one from a library. Apache Commons has Fraction and BigFraction, though their documentation seems a little sketchy.

Java is performing integer division because your denominator is an integer.
Try the following:
double quarter = 1 / 4.0;
Or:
double quarter = 1 / (double) 4;

The reason you're getting 0.0 is because the division is done as an integer division and then the result is converted to float. Try this, for example: double quarter = 1.0/4.0; - you should get (pretty much) the expected result.
However, depending on your requirements, this may not be the best way to deal with the problem. For example, you can't store 1/3 in a decimal. The perfect way would be to store simple fraction as a pair of integers. You can create a class for it (with some arithmetic methods) or start by using a simple array. It all depends on your needs.

How can I handle precision error with float in Java?

I'm wondering what the best way to fix precision errors is in Java. As you can see in the following example, there are precision errors:
class FloatTest
{
public static void main(String[] args)
{
Float number1 = 1.89f;
for(int i = 11; i < 800; i*=2)
{
System.out.println("loop value: " + i);
System.out.println(i*number1);
System.out.println("");
}
}
}
The result displayed is:
loop value: 11
20.789999
loop value: 22
41.579998
loop value: 44
83.159996
loop value: 88
166.31999
loop value: 176
332.63998
loop value: 352
665.27997
loop value: 704
1330.5599
Also, if someone can explain why it only does it starting at 11 and doubling the value every time. I think all other values (or many of them at least) displayed the correct result.
Problems like this have caused me headache in the past and I usually use number formatters or put them into a String.
Edit: As people have mentioned, I could use a double, but after trying it, it seems that 1.89 as a double times 792 still outputs an error (the output is 1496.8799999999999).
I guess I'll try the other solutions such as BigDecimal

If you really care about precision, you should use BigDecimal
https://docs.oracle.com/javase/8/docs/api/java/math/BigDecimal.html
https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/math/BigDecimal.html

The problem is not with Java but with the good standard float's (http://en.wikipedia.org/wiki/IEEE_floating-point_standard).
You can either:
use Double and have a bit more precision (but not perfect of course, it also has limited precision)
use a arbitrary-precision-library
use numerically stable algorithms and truncate/round digits of which you are not sure they are correct (you can calculate numeric precision of operations)

When you print the result of a double operation you need to use appropriate rounding.
System.out.printf("%.2f%n", 1.89 * 792);
prints
1496.88
If you want to round the result to a precision, you can use rounding.
double d = 1.89 * 792;
d = Math.round(d * 100) / 100.0;
System.out.println(d);
prints
1496.88
However if you see below, this prints as expected, as there is a small amount of implied rounding.
It worth nothing that (double) 1.89 is not exactly 1.89 It is a close approximation.
new BigDecimal(double) converts the exact value of double without any implied rounding. It can be useful in finding the exact value of a double.
System.out.println(new BigDecimal(1.89));
System.out.println(new BigDecimal(1496.88));
prints
1.8899999999999999023003738329862244427204132080078125
1496.8800000000001091393642127513885498046875

Most of your question has been pretty well covered, though you might still benefit from reading the [floating-point] tag wiki to understand why the other answers work.
However, nobody has addressed "why it only does it starting at 11 and doubling the value every time," so here's the answer to that:
for(int i = 11; i < 800; i*=2)
╚═══╤════╝ ╚╤═╝
│ └───── "double the value every time"
│
└───── "start at 11"

You could use doubles instead of floats

If you really need arbitrary precision, use BigDecimal.

first of Float is the wrapper class for the primitive float
and doubles have more precision
but if you only want to calculate down to the second digit (for monetary purposes for example) use an integer (as if you are using cents as unit) and add some scaling logic when you are multiplying/dividing
or if you need arbitrary precision use BigDecimal

If precision is vital, you should use BigDecimal to make sure that the required precision remains. When you instantiate the calculation, remember to use strings to instantiate the values instead of doubles.

I never had a problem with simple arithmetic precision in either Basic, Visual Basic, FORTRAN, ALGOL or other "primitive" languages. It is beyond comprehension that JAVA can't do simple arithmetic without introducing errors. I need just two digits to the right of the decimal point for doing some accounting. Using Float subtracting 1000 from 1355.65 I get 355.650002! In order to get around this ridiculous error I have implemented a simple solution. I process my input by separating the values on each side of the decimal point as character, convert each to integers, multiply each by 1000 and add the two back together as integers. Ridiculous but there are no errors introduced by the poor JAVA algorithms.