Arithmetic limitations
Expect the unexpected:
Figure 103. Strange things I 
byte count = 91; // o.K.
int i = 91;
byte count2 = i; // Compile error: Incompatible types
// Required: byte Found: int
byte points = 130; // Compile error: Incompatible types
// Required: byte Found: intFigure 104. Strange things II
final int i = 91;
byte count = i; // o.K.Figure 105. Arithmetic overflow pitfalls
int count = 2147483647;
int points = 2147483647;
int sum = count + points; ❶
System.out.println("Sum = " + sum);Result:
Sum = -2 |
01111111_11111111_11111111_11111111 + 01111111_11111111_11111111_11111111 _____________________________________ 11111111_11111111_11111111_11111110 |
|
This exceeds the four byte limit of int variables. Thus the
leading “1” will be discarded leaving us with a result
of |
Why not generally using float
/ double
in favour of seemingly more limited byte,
short,
int,
long
for arithmetics?
Figure 106. Limited precision
float float2Power31 = Integer.MAX_VALUE + 1f; // 2^31
float floatDoubleMAX_VALUE = 2 * float2Power31 * float2Power31 - 1f; // 2^63 - 1
System.out.format( " Float value: %f\n", floatDoubleMAX_VALUE);
System.out.println("Expected value: " + Long.MAX_VALUE);Result:
Float value: 9223372036854776000.000000
Expected value: 9223372036854775807
Difference: 193Figure 107. Nearest
float to 2.1 DecimalFormat df = new DecimalFormat("#.###############"); //Print 15 floating point digits
System.out.println(df.format(Float.intBitsToFloat(0b0_10000000_00001100110011001100110 )));
System.out.println(df.format(Float.intBitsToFloat(0b0_10000000_00001100110011001100111 ))); |
2.099999904632568 2.100000143051147 |
Figure 108.
FloatConverter 
