# 1.6. Casting and Ranges of Values¶

In Java, **type casting** is used to convert values from one type to another. By
**casting** we don’t mean something to do with fishing, but it is a similar idea
to casting a bronze, without needing to heat anything to 913 degrees Celsius.
But like molten bronze is reshaped by melting it and pouring it into a mold, our
data is reshaped via a **cast** operator. In Java when you cast you are changing
the “shape” (or type) of the value.

The **cast operator**, which looks like `(int)`

and `(double)`

placed before
any expression (a literal, a number, a variable, or more complex expression in
parentheses) produces a value of the given type by converting the value of the
original expression to the new type.

For example, `(double) 1 / 3`

will evaluate to a `double`

value instead of an
`int`

. Run this code to find how Java handles division and what casting can do
to the results. Notice what happens when you divide an `int`

by an `int`

or
an `int`

by a `double`

or an `int`

cast to a `double`

divided by an
`int`

.

What happens when you divide an int by an int or with a double operand or with the type cast (double) on one of the operands?

When Java divides two `int`

s, it produces an `int`

result by truncating
the actual mathematical result, removing anything after the decimal point. Thus
`9 / 10`

evaluates to `0`

, not `0.9`

. It also does not evaluate to `1`

;
truncating is not the same as rounding!

But in any expression involving a `double`

, the `double`

is “contagious” and
will cause the value of that expression to also be a `double`

. Thus the
expression `9.0 / 10`

is evaluated as if it had been written `9.0 / 10.0`

and
produces the `double`

value `0.9`

.

Casting an `int`

to `double`

, as shown in the code above, produces a
`double`

value which will then causes any expression it is part of to produce
a `double`

. This is especially useful when you have `int`

variables that
you want to do non-integer division with:

```
int total; // a variable containing the sum of a bunch of ints
int count; // the number of ints that went into total
// Compute the average of the bunch of ints summed into total.
double average = (double) total / count;
```

A conversion from `int`

to `double`

is called a **widening conversion**
because a `double`

can represent any `int`

value but not vice versa; thus a
`double`

is considered a wider data type than an `int`

.

Note

`int`

s in Java are always 32-bit signed values which mean they can
represent values from \(-2^{31}\) to \(2^{31} - 1\), inclusive, while
the range of consecutive integer values that can be represented by a double
is from \(-2^{53}\) to \(2^{53}\), inclusive. (A `double`

can also
represent much larger values but with limited precision.) You can refer to
the minimum and maximum `int`

values with the constants
`Integer.MIN_VALUE`

and `Integer.MAX_VALUE`

.

Values of type `double`

in the range that can be represented by an `int`

can
be rounded to the nearest `int`

by adding or subtracting 0.5 and then casting
the result with `(int)`

:

```
double number; // positive value from somewhere
double negNumber; // negative value from somewhere
int nearestInt = (int)(number + 0.5);
int nearestNegInt = (int)(negNumber – 0.5);
```

For example, if you divide `7.0 / 4.0`

you get `1.75`

. If you cast that to
an `int`

, it will be truncated to `1`

. However if we want to round a
`double`

rather than truncating it, adding `0.5`

will produce a number that
is above the next integer value if the decimal part is greater than `0.5`

, as
it is here. Then casting *that* value to an `int`

will truncate down. So in
this case `1.75 + 0.5`

gives us `2.25`

which is then truncated to `2`

. On
the other hand adding `0.5`

to the result of evaluating `5.0 / 4.2`

, namely
`1.25`

, only gets us to `1.75`

which truncates back to `1`

which is the
nearest integer to `1.25`

.

Run the code below to see how the formula of adding or subtracting .5 and then casting with (int) rounds a positive or negative double number to the closest int.

What happens to repeating decimal numbers like 3.333333…? Java limits the number of digits you can save for any `double`

number to about 14-15 digits. You should be aware that the accuracy of any calculation on a computer is limited by the fact that computers can only hold a limited number of digits.

For example, int values are stored in 4 bytes of memory. There is an
`Integer.MAX_VALUE`

defined as 2147483647 and an `Integer.MIN_VALUE`

defined
as -2147483648. If you try to store any number larger or smaller than these
numbers in an int variable, it will result in an **integer overflow** where an
incorrect value could be stored. Try it below.

Try the code below to see two integer overflows for a positive and negative number. An int cannot hold that many digits! Fix the integer overflow by deleting the last 0 in the numbers to store less digits.

Although it’s not on the AP exam, you can format long decimal numbers to just show 2 digits after the decimal point with the following code:

Run the code below to see how a decimal number can be formatted to show 2 digits after the decimal point.

- true
- Did you try this out in Active Code? Does it work that way?
- false
- Java throws away any values after the decimal point if you do integer division. It does not round up automatically.

1-6-5: True or false: Java rounds up automatically when you do integer division.

- true
- Try casting to int instead of double. What does that do?
- false
- Casting results in the type that you cast to. However, if you can't really cast the value to the specified type then you will get an error.

1-6-6: True or false: casting always results in a double type.

- (double) (total / 3);
- This does integer division before casting the result to double so it loses the fractional part.
- total / 3;
- When you divide an integer by an integer you get an integer result and lose the fractional part.
- (double) total / 3;
- This will convert total to a double value and then divide by 3 to return a double result.

1-6-7: Which of the following returns the correct average for a total that is the sum of 3 int values?

## 1.6.1. Programming Challenge : Average 3 Numbers¶

This would be a good project to work together in pairs, and switch drivers (who has control of the keyboard in pair programming) after every line of code. In the code below, type in three made up int grades and then sum and average them. Use casting to report the result as a double. For example, if the grades are 90, 100, and 94, the sum of the three numbers is 90 + 100 + 94 = 284, and the average is the sum 284 divided by 3 which casted to a double is 94.666667. You should use your variables instead of the numbers in your formulas. Follow the pseudocode below.

Type in three made up int grades and then sum and average them. Use type casting to report the result as a double. If you do this challenge on replit.com (see template and links below), please paste your repl link here to turn it in.

Your teacher may suggest that you use a Java IDE like replit for this challenge so that you can use input to get these values using the Scanner class. Here is a repl template that you can use to get started if you want to try the challenge with input.

## 1.6.2. Bonus Challenge : Unicode¶

If you get done early with the previous challenge, here’s something else fun you can do in Java, although it’s not covered in the AP exam.

Java was one of the first programming languages to use Unicode for its
characters rather than ASCII. While ASCII could represent 128 characters which
was plenty for English, Unicode is an international standard that tries to
assign a number (which they like to call a “codepoint”) to every character in
every language. Unicode codepoints are traditionally represented in hex code (a
base 16 code that uses the digits 0-9 and the letters A-F for 10-15), so you
might see things like `U+1F600`

. But they’re just numbers. That last one is
the same as `128512`

.

When Java was released in an 1996, Unicode had been around for five years and
the Unicode people had declared they would only ever need 2^{16} or 65,536
code points to represent all the characters used in the world. So Java included
a `char`

data type that can hold exactly 2^{16} values. Then, seven
months later, the Unicode folks, said, “Ooops, that’s not enough”, and extended
their system to its current size of 1,112,064 possible codepoints. (As of
September 2022, 149,186 have actually been used.)

That made `char`

kind of obsolete. But while not every Unicode codepoint can
be represented in a Java `char`

, you can use an `int`

to represent a
codepoint and the method `Character.toString`

to translate an `int`

into a
`String`

containing the character for that codepoint. (You might see older
Java code that casts numbers to `char`

; for many codepoints that will work but
not on more recently added codepoints including, critically those for Emoji. 😞
So better to use `Character.toString`

and ignore `char`

.)

Try the following program which prints out an English “A”, a Chinese character, and an emoji. Then look up other characters at this Unicode Lookup site and change the code to print them out. (Use the Dec column in site to get the decimal number.) Can you print out letters from 3 different languages?

Can you print out a letter from 3 different languages using this Unicode Lookup site?

## 1.6.3. Summary¶

**Type casting**is used to convert value from one type to another.The casting operators

`(int)`

and`(double)`

can be used to create a temporary value converted to a different data type.Casting a

`double`

value to an`int`

causes the digits to the right of the decimal point to be truncated (cut off and thrown away).In expressions involving

`double`

s, the`double`

values are contagious, causing`int`

s in the expression to be automatically converted to the equivalent`double`

value so the result of the expression can be computed as a`double`

.Values of type

`double`

can be rounded to the nearest integer by (int)(x + 0.5) or (int)(x – 0.5) for negative numbers.Integer values in Java are represented by values of type

`int`

, which are stored using a finite amount (4 bytes) of memory. Therefore, an int value must be in the range from`Integer.MIN_VALUE`

to`Integer.MAX_VALUE`

, inclusive.If an expression would evaluate to an int value outside of the allowed range, an integer overflow occurs. This could result in an incorrect value within the allowed range.