Conversions

Figure 105. Widening from byte literal to short

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Figure 106. Narrowing from int literal to char variable

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Figure 107. A widening «ladder»

byte   b = 42;  // Narrowing: constant int literal to byte
short  s = b;   // Widening
int    i = s;   // Widening
long   l = i;   // Widening
float  f = l;   // Widening
double d = f;   // Widening

Figure 108. A narrowing «ladder»

double d = 14.23;
float  f  = (float) d;  // Narrowing
long   l  = (long)  f;  // Narrowing
int    i  = (int)   l;  // Narrowing
short  s  = (short) i;  // Narrowing
byte   b  = (byte)  s;  // Narrowing

exercise No. 27

`int` and `char`

Explain the following output:

Code	Output
`char a = 0; a -= 1; int i = a; System.out.println(i);`	65535

Why do we see a result of 65535 rather than -1?

Tip

Read about the a char's value range.

In contrast to byte, short and int a two byte char represents no negative but only positive values and zero:

decimal	binary
0	`0000 0000 0000 0000`
1	`0000 0000 0000 0001`
2	`0000 0000 0000 0010`
...
65535 ( $2^{16} - 1$ )	`1111 1111 1111 1111`

Like with other integer types the above table behaves in a cyclic way with respect to additions at its top and and subtractions at its bottom:

Code	Output
`char a = 65535; a += 1; int i = a; System.out.println(i);`	0

On machine level this may be conceived as an (incomplete) subtract operation:

 0000 0000 0000 0000
-0000 0000 0000 0001
--------------------
 1111 1111 1111 1111

exercise No. 28

`float` vs. `double`

We consider:

Code	Output
`System.out.println( 3.14 ); System.out.println( 3.14f ); System.out.println( 3.14d );`	3.14 3.14 3.14

There seems to be no difference between the three literals 3.14, 3.14f and 3.14d. Is this actually true?

Tip

Read the section about floating point literals in Java™. Write code exhibiting possible differences.

The System.out.println( 3.14f ) statement's output is actually truncated with respect to output precision. Forcing 16 fractional digits to become visible reads:

Code	Output
`System.out.println(new DecimalFormat( "#0.0000000000000000").format(3.14f));`	3.1400001049041750

The value 3.1400001049041750 is the closest possible approximation to 3.14 when using a 4-byte IEEE float. A 3.14f double literal won't be exact either but doubles the number of representing bytes to 8 thereby enhancing its representational precision substantially. This causes unexpected results:

Code	Output
System.out.println(3.14f - 3.14d);	1.0490417468034252E-7

This difference is a result of 3.14d providing four additional bytes of precision therefore matching the exact value of $\frac{314}{100}$ better than 3.14f.

Regarding types we have:

Literal	Comment
`3.14f`	4-byte `float` literal
`3.14d`	8-byte `double` literal
`3.14`	Equivalent to `3.14d`

Assignments to variables of type double are thus always guaranteed to work:

Code	Comment
`double d = 3.14f;`	o.K., assigning `float` to `double` (widening)
`double d = 3.14d;`	o.K., assigning `double` to `double`
`double d = 3.14;`	o.K., assigning `double` to `double` as well.

Assignments to variables of type float may fail:

Code	Comment
`float f = 3.14f;`	o.K., assigning `float` to `float`
float f = 3.14d;	Error, assigning `double` to `float`
float f = 3.14;	Error, assigning (implicit) `double` to `float` as well.

exercise No. 29

`int` to `char` narrowing problems

Reconsidering Figure 106, “Narrowing from int literal to char variable ” we observe the following two related snippets yielding compile time errors

```
int i = 65;
char c = i;
```
On contrary the following code compiles well:
```
char c = 65;
```
```
char c = 66200;
```
On contrary the following code compiles well:
```
char c = 64200;
```

Explain these errors and their underlying reasons and provide a solution if possible.

Tip

Which data types are involved? Think about narrowing conversions and type casting.

Assigning an int to a char variable effectively narrows from four to two bytes and is thus prohibited. A fix requires an explicit type cast:
```
int i = 65;
char c = (char) i;
```
Since 65 fits well into a char being limited by $2^{16} - 1$ our cast will not have any negative impact.
We consider the binary representations of 66200 and 64200:
```
System.out.println(Integer.toBinaryString(66200)); // Yields 1_00000010_10011000
System.out.println(Integer.toBinaryString(64200)); // Yields   11111010_11001000
```
Thus 64200 fits into a two byte char whereas 66200 being larger than $2^{16} - 1$ does not.

exercise No. 30

Get a `byte` from 139

Consider:

int i = 139;
byte b = (byte) i;
System.out.println(b);

Explain in detail why execution results in a value of -117.

A four byte int representation of 139 reads 00000000_00000000_00000000_10001011. The cast b = (byte) i will strip the leading three bytes leaving us with b containing 10001011.

Since byte values in Java™ are being represented as signed values in two-complement notation this equals decimal -117.

exercise No. 31

Ariane, I miss you!

Reconsidering the Ariane 5 maiden flight crash read the comment buried in the solution of Inventing tinyint. . Try to mimic a code portion in Java™ showing the catastrophic error.

Start with a double variable using a value being suitable to be assigned to a short variable using a cast (narrowing).

Then in a second step raise this value breaking your short variable's upper limit.

We start from:

double level = 2331.12; // smaller than 32767 == Short.MAX_VALUE

short value = (short) level;

System.out.println(value);

Execution yields an expected integer output of 2331. However increasing our level variable's value from 2331.12 to 42331.12 yields an output of -23205 due to an overflow.

exercise No. 32

Reducing `long` to `int` (difficult)

For changing a map's scale from fine to coarse Joe programmer intends to map positive long values to int values. This requires scaling down “half” the long data type's range $[0, 2^{63} - 1]$ to the int's range of $[0, 2^{31} - 1]$ :

From		To
`long`	remark	`int`	remark
0		0
1		0
...		0
4294967295	$2^{32} - 1$	0
4294967296	$2^{32}$	1
4294967297	$2^{32} + 1$	1
...
9223372036854775806	$2^{63} - 2$ or `Long.MAX_VALUE` - 1	2147483647	$2^{31} - 1$ or `Integer.MAX_VALUE`
9223372036854775807	$2^{63} - 1$ or `Long.MAX_VALUE`	2147483647	$2^{31} - 1$ or `Integer.MAX_VALUE`

Joe's idea is dividing long values by $2^{32}$ . As a result a long value of $- 2^{63}$ will be reduced to the intended value of $- 2^{31}$ . Since $2^{32}$ seems to be equal to 2 * (Integer.MAX_VALUE + 1)) (why?) Joe's first attempt reads:

final long longValue = 2147483648L;
final int reducedValue = (int) (longValue / (2 * (Integer.MAX_VALUE + 1)));
System.out.println(reducedValue);

Unfortunately the results are not promising. This code merely results in a runtime error:

/usr/lib/jvm/java-8-oracle/bin/java ...
Exception in thread "main" java.lang.ArithmeticException: / by zero
	at qq.App.main(App.java:27)

Process finished with exit code 1

Explain the underlying problem and correct Joe's error.

Tip

It may be helpful thinking of a smaller example before. Consider two hypothetic signed integer types “TinyLong” and “TinyInt” of four and two bits respectively. The corresponding mapping will be:

“TinyLong”, n = 4	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6	7
“TinyInt”, n = 2	-2				-1				0				1

Joe's intention with respect to our toy example types implies dividing “TinyLong” values by $2^{2}$ (truncating). This indeed yields the desired result for non-negative values. So why does Joe encounter a division by zero runtime exception when executing longValue / (2 * (Integer.MAX_VALUE + 1))?

Unfortunately Joe's implementation is seriously flawed for even two reasons:

The constant Integer.MAX_VALUE already suggests we will not be able to increase its value while staying as an int. The expression Integer.MAX_VALUE + 1 will be evaluated using int rather than long arithmetic returning:
```
  01111111_11111111_11111111_11111111
+ 00000000_00000000_00000000_00000001
_____________________________________
  10000000_00000000_00000000_00000000
```
This is the binary representation of the unintended result Integer.MIN_VALUE due to an arithmetic overflow. The expression 2 * (Integer.MAX_VALUE + 1) then gives rise to a second overflow error:
```
  10000000_00000000_00000000_00000000
+ 10000000_00000000_00000000_00000000
_____________________________________
  00000000_00000000_00000000_00000000
```

Both errors combined surprisingly result in a value of 0 explaining the “division by zero” error message. There are two possible solutions:

(int) (longValue / (2L * (Integer.MAX_VALUE + 1L))): Introducing 2L or 1L (one is sufficient) in favour of simply using 2 and 1 turns both addition and multiplication into operations involving at least one long argument. Thus for both operations the Java™ runtime will use long arithmetic returning the desired “reducing” factor of $2^{32}$ of type long.
(int) (longValue / 2 / (Integer.MAX_VALUE + 1L)): Same result as before.

Note

This time the expression starts with longValue / 2 ... Since the variable longValue is of type long the expression longValue / 2 will be evaluated by the Java™ runtime using long arithmetics. The result will subsequently be divided by Integer.MAX_VALUE + 1L again using long arithmetic.

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Language Fundamentals

Conversions

int and char

Tip

float vs. double

Tip

int to char narrowing problems

Tip

Get a byte from 139