[java] Int division: Why is the result of 1/3 == 0?

I was writing this code:

public static void main(String[] args) {
    double g = 1 / 3;
    System.out.printf("%.2f", g);
}

The result is 0. Why is this, and how do I solve this problem?

Because it treats 1 and 3 as integers, therefore rounding the result down to 0, so that it is an integer.

To get the result you are looking for, explicitly tell java that the numbers are doubles like so:

double g = 1.0/3.0;

The easiest solution is to just do this

double g = (double) 1 / 3;

What this does, since you didn't enter 1.0 / 3.0, is let you manually convert it to data type double since Java assumed it was Integer division, and it would do it even if it meant narrowing the conversion. This is what is called a cast operator. Here we cast only one operand, and this is enough to avoid integer division (rounding towards zero)

Many others have failed to point out the real issue:

An operation on only integers casts the result of the operation to an integer.

This necessarily means that floating point results, that could be displayed as an integer, will be truncated (lop off the decimal part).

What is casting (typecasting / type conversion) you ask?

It varies on the implementation of the language, but Wikipedia has a fairly comprehensive view, and it does talk about coercion as well, which is a pivotal piece of information in answering your question.

http://en.wikipedia.org/wiki/Type_conversion

Make the 1 a float and float division will be used

public static void main(String d[]){
    double g=1f/3;
    System.out.printf("%.2f",g);
}

Try this out:

public static void main(String[] args) {
    double a = 1.0;
    double b = 3.0;
    double g = a / b;
    System.out.printf(""+ g);
}

Because you are doing integer division.

As @Noldorin says, if both operators are integers, then integer division is used.

The result 0.33333333 can't be represented as an integer, therefore only the integer part (0) is assigned to the result.

If any of the operators is a double / float, then floating point arithmetic will take place. But you'll have the same problem if you do that:

int n = 1.0 / 3.0;

you should use

double g=1.0/3;

or

double g=1/3.0;

Integer division returns integer.

I did this.

double g = 1.0/3.0;
System.out.printf("%gf", g);

Use .0 while doing double calculations or else Java will assume you are using Integers. If a Calculation uses any amount of double values, then the output will be a double value. If the are all Integers, then the output will be an Integer.

The result is 0. Why is this, and how do I solve this problem?

TL;DR

You can solve it by doing:

double g = 1.0/3.0; 

or

double g = 1.0/3; 

or

double g = 1/3.0; 

or

double g = (double) 1 / 3;

The last of these options is required when you are using variables e.g. int a = 1, b = 3; double g = (double) a / b;.

A more completed answer

double g = 1 / 3;

This result in 0 because

• first the dividend < divisor;
• both variables are of type int therefore resulting in int (5.6.2. JLS) which naturally cannot represent the a floating point value such as 0.333333...
• "Integer division rounds toward 0." 15.17.2 JLS

Why double g = 1.0/3.0; and double g = ((double) 1) / 3; work?

From Chapter 5. Conversions and Promotions one can read:

One conversion context is the operand of a numeric operator such as + or *. The conversion process for such operands is called numeric promotion. Promotion is special in that, in the case of binary operators, the conversion chosen for one operand may depend in part on the type of the other operand expression.

and 5.6.2. Binary Numeric Promotion

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order:

If any operand is of a reference type, it is subjected to unboxing conversion (§5.1.8).

Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

If either operand is of type double, the other is converted to double.

Otherwise, if either operand is of type float, the other is converted to float.

Otherwise, if either operand is of type long, the other is converted to long.

Otherwise, both operands are converted to type int.

public static void main(String[] args) {
    double g = 1 / 3;
    System.out.printf("%.2f", g);
}

Since both 1 and 3 are ints the result not rounded but it's truncated. So you ignore fractions and take only wholes.

To avoid this have at least one of your numbers 1 or 3 as a decimal form 1.0 and/or 3.0.

(1/3) means Integer division, thats why you can not get decimal value from this division. To solve this problem use:

public static void main(String[] args) {
        double g = 1.0 / 3;
        System.out.printf("%.2f", g);
    }

1/3 uses integer division as both sides are integers.

You need at least one of them to be float or double.

If you are entering the values in the source code like your question, you can do 1.0/3 ; the 1.0 is a double.

If you get the values from elsewhere you can use (double) to turn the int into a double.

int x = ...;
int y = ...;
double value = ((double) x) / y;

The conversion in JAVA is quite simple but need some understanding. As explain in the JLS for integer operations:

If an integer operator other than a shift operator has at least one operand of type long, then the operation is carried out using 64-bit precision, and the result of the numerical operator is of type long. If the other operand is not long, it is first widened (§5.1.5) to type long by numeric promotion (§5.6).

And an example is always the best way to translate the JLS ;)

int + long -> long
int(1) + long(2) + int(3) -> long(1+2) + long(3)

Otherwise, the operation is carried out using 32-bit precision, and the result of the numerical operator is of type int. If either operand is not an int, it is first widened to type int by numeric promotion.

short + int -> int + int -> int

A small example using Eclipse to show that even an addition of two shorts will not be that easy :

short s = 1;
s = s + s; <- Compiling error

//possible loss of precision
//  required: short
//  found:    int

This will required a casting with a possible loss of precision.

The same is true for the floating point operators

If at least one of the operands to a numerical operator is of type double, then the operation is carried out using 64-bit floating-point arithmetic, and the result of the numerical operator is a value of type double. If the other operand is not a double, it is first widened (§5.1.5) to type double by numeric promotion (§5.6).

So the promotion is done on the float into double.

And the mix of both integer and floating value result in floating values as said

If at least one of the operands to a binary operator is of floating-point type, then the operation is a floating-point operation, even if the other is integral.

This is true for binary operators but not for "Assignment Operators" like +=

A simple working example is enough to prove this

int i = 1;
i += 1.5f;

The reason is that there is an implicit cast done here, this will be execute like

i = (int) i + 1.5f
i = (int) 2.5f
i = 2

1 and 3 are integer contants and so Java does an integer division which's result is 0. If you want to write double constants you have to write 1.0 and 3.0.

Explicitly cast it as a double

double g = 1.0/3.0

This happens because Java uses the integer division operation for 1 and 3 since you entered them as integer constants.

Do "double g=1.0/3.0;" instead.