Nov 15, 2000 compiled on — Thursday August 25, 2016 at 05:31 AM

1 Java primitive types sizes

2 Maximum value in signed and unsigned integers

3 Some bits table

4 Power of 2 table

5 Float and Double in Java

5.1 How to read a floating point?

6 References

2 Maximum value in signed and unsigned integers

3 Some bits table

4 Power of 2 table

5 Float and Double in Java

5.1 How to read a floating point?

6 References

type | size in bytes |

byte | 1 |

short | 2 |

int | 4 |

long | 8 |

float | 4 (IEEE 754) |

double | 8 (IEEE 754) |

Signed integer table

number of bits | Java type | range | range in base 10 |

8 | byte | ||

16 | short | ||

32 | int | ||

64 | long | ||

number of bits | Java type | range | range in HEX |

8 | byte | 7F | |

16 | short | 7F FF | |

32 | int | 7F FF FF FF | |

64 | long | 7F FF FF FF FF FF FF FF | |

Unsigned integer table

number of bits | Java type | range | range in base 10 |

8 | byte | ||

16 | short | ||

32 | int | ||

64 | long | ||

number of bits | Java type | range | range in HEX |

8 | byte | FF | |

16 | short | FF FF | |

32 | int | FF FF FF FF | |

64 | long | FF FF FF FF FF FF FF FF | |

The max value that can be obtained using bits is found by using the formula , this assume unsignd values.

bit pattern | base 10 | Hex |

0 | 0 | 0 |

1 | 1 | 1 |

10 | 2 | 2 |

11 | 3 | 3 |

100 | 4 | 4 |

101 | 5 | 5 |

110 | 6 | 6 |

111 | 7 | 7 |

1000 | 8 | 8 |

1001 | 9 | 9 |

1010 | 10 | A |

1011 | 11 | B |

1100 | 12 | C |

1101 | 13 | D |

1110 | 14 | E |

1111 | 15 | F |

1 0000 | 16 | 10 |

1 0001 | 17 | 11 |

1 0010 | 18 | 12 |

1 0011 | 19 | 13 |

1 0100 | 20 | 14 |

1 0101 | 21 | 15 |

1 0110 | 22 | 16 |

1 0111 | 23 | 17 |

1 1000 | 24 | 18 |

1 1001 | 25 | 19 |

1 1010 | 26 | 1A |

1 1011 | 27 | 1B |

1 1100 | 28 | 1C |

1 1101 | 29 | 1D |

1 1110 | 30 | 1E |

1 1111 | 31 | 1F |

10 0000 | 32 | 20 |

0111 1111 | 127 | 7F |

10000000 | 128 | 80 |

11111111 | 255 | FF |

1 00000000 | 256 | 1 00 |

1111 11111111 | F FF | |

11111111 11111111 | FF FF | |

1111 11111111 11111111 | F FF FF | |

11111111 11111111 11111111 | FF FF FF | |

1111 11111111 11111111 11111111 | F FF FF FF | |

11111111 11111111 11111111 11111111 | FF FF FF FF | |

So, 16 bits needs 5 digits in base 10 to represent it.

32 bits needs 10 digits in base 10 to represent it.

64 bits needs 20 digits in base 10 to represent it.

So, it looks like the number of digits in base 10 to represent a bit pattern of length is

So 128 bits will require about 42 digits in base 10 to represent externally.

power of two | base 2 | base 10 | Hex |

1 | 1 | 1 | |

01 | 2 | 2 | |

100 | 4 | 4 | |

1000 | 8 | 8 | |

1 0000 | 16 | 10 | |

10 0000 | 32 | 20 | |

100 0000 | 64 | 40 | |

1000 0000 | 128 | 80 | |

1 0000 0000 | 256 | 1 00 | |

10 0000 0000 | 512 | 2 00 | |

… | (1K) | 4 00 | |

8 00 | |||

10 00 | |||

20 00 | |||

40 00 | |||

80 00 | |||

1 00 00 | |||

2 00 00 | |||

4 00 00 | |||

8 00 00 | |||

(1 MB) | 10 00 00 | ||

20 00 00 | |||

40 00 00 | |||

80 00 00 | |||

1 00 00 00 | |||

2 00 00 00 | |||

4 00 00 00 | |||

8 00 00 00 | |||

10 00 00 00 | |||

20 00 00 00 | |||

(1 GB) | 40 00 00 00 | ||

80 00 00 00 | |||

1 00 00 00 00 | |||

2 00 00 00 00 | |||

4 00 00 00 00 | |||

8 00 00 00 00 | |||

10 00 00 00 00 | |||

20 00 00 00 00 | |||

40 00 00 00 00 | |||

80 00 00 00 00 | |||

(1 tera) | 1 00 00 00 00 00 | ||

2 00 00 00 00 00 | |||

4 00 00 00 00 00 | |||

8 00 00 00 00 00 | |||

10 00 00 00 00 00 | |||

20 00 00 00 00 00 | |||

40 00 00 00 00 00 | |||

power of two | base 2 | base 10 | Hex |

100000… | 80 00 00 00 00 00 | ||

1 00 00 00 00 00 00 | |||

2 00 00 00 00 00 00 | |||

4 00 00 00 00 00 00 | |||

8 00 00 00 00 00 00 | |||

10 00 00 00 00 00 00 | |||

20 00 00 00 00 00 00 | |||

40 00 00 00 00 00 00 | |||

80 00 00 00 00 00 00 | |||

1 00 00 00 00 00 00 00 | |||

2 00 00 00 00 00 00 00 | |||

4 00 00 00 00 00 00 00 | |||

8 00 00 00 00 00 00 00 | |||

10 00 00 00 00 00 00 00 | |||

20 00 00 00 00 00 00 00 | |||

40 00 00 00 00 00 00 00 | |||

80 00 00 00 00 00 00 00 | |||

1 00 00 00 00 00 00 00 00 | |||

Java uses IEEE 754.

A number such as is expressed as or .

In floating point, the second form above is used. i.e. base 2 is used for the exponent.

The sign uses 1 bit. 0 for positive and 1 for negative. The exponent uses the next 8 bits (biased by 127), and the exponent uses the next 23 bits.

In Java, a float uses IEEE 754. The following explains how float and double represented in Java.

So, from the above, a float in IEEE 754 is in the range

In Java a double is expressed as

So, from the above, a double in IEEE 754 is in the range

Given this example:

11000011100101100000000000000000

The above is binary representation of single precision floating point (32 bit).

Reading from the left most bit (bit 31) to the right most bit (bit 0).

bit 31 is 1, so this is a negative number. bits 30 …23 is the exponent, which is 10000111 or 135. But since the exponent is biased by 127, it is actually 8, so now we have the exponent part which is . Next is bits 22 …0, which is 00101100000000000000000, since there is an implied 1, the above can be re-written as 1.00101100000000000000000, which is read as follows:

which is

Hence the final number is .

The above implies that a number that be can't be expressed as sum of power of 2, can't be represented exactly in a floating point. Since a float is represented as , assume , then the accuracy of a float goes like this: or ,

So, a number such as can't be exactly expressed in floating point ! because the value can't be expressed as a sum of power of 2.

The greatest number that has an exact IEEE single-precision representation is 340282346638528859811704183484516925440.0 , This is 40 digits number, which is represented by

The Java programing language specifications.

http://www.math.grin.edu/~stone/courses/fundamentals/IEEE-reals.html