Nov 15, 2000 Compiled on May 20, 2020 at 5:41am

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 | \(2^7-1 \ldots -2^7\) | \(127 \ldots -128\) |

16 | short | \(2^{15}-1 \ldots -2^{15}\) | \(32,767 \ldots -32,768\) |

32 | int | \(2^{31}-1 \ldots -2^{31}\) | \(2,147,483,647 \ldots -2,147,483,648\) |

64 | long | \(2^{63}-1 \ldots -2^{63}\) | \(9,223,372,036,854,775,807 \ldots -9,223,372,036,854,775,808\) |

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

8 | byte | \(2^7-1 \ldots -2^7\) | 7F \(\ldots -80\) |

16 | short | \(2^{15}-1 \ldots -2^{15}\) | 7F FF \(\ldots -80 00\) |

32 | int | \(2^{31}-1 \ldots -2^{31}\) | 7F FF FF FF \(\ldots -80 00 00 00\) |

64 | long | \(2^{63}-1 \ldots -2^{63}\) | 7F FF FF FF FF FF FF FF \(\ldots -80 00 00 00 00 00 00 00\) |

Unsigned integer table

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

8 | byte | \(2^8-1 \ldots 0\) | \(255 \ldots 0\) |

16 | short | \(2^{16}-1 \ldots 0\) | \(65,535 \ldots 0\) |

32 | int | \(2^{32}-1 \ldots 0\) | \(4,294,967,295 \ldots 0\) |

64 | long | \(2^{64}-1 \ldots 0\) | \(18,446,744,073,709,551,615 \ldots 0\) |

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

8 | byte | \(2^8-1 \ldots 0\) | FF \(\ldots 00\) |

16 | short | \(2^{16}-1 \ldots 0\) | FF FF \(\ldots 00 00\) |

32 | int | \(2^{32}-1 \ldots 0\) | FF FF FF FF \(\ldots 00 00 00 00\) |

64 | long | \(2^{64}-1 \ldots 0\) | FF FF FF FF FF FF FF FF \(\ldots 00 00 00 00 00 00 00 00\) |

The max value that can be obtained using \(n\) bits is found by using the formula \(2^n-1\), 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 | \(4,095\) | F FF |

11111111 11111111 | \(65,535\) | FF FF |

1111 11111111 11111111 | \(1,048,575\) | F FF FF |

11111111 11111111 11111111 | \(16,777,215\) | FF FF FF |

1111 11111111 11111111 11111111 | \(268,435,455\) | F FF FF FF |

11111111 11111111 11111111 11111111 | \(4,294,967,295\) | 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 \(n\) is \((1/3) n\)

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

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

\(2^0\) | 1 | 1 | 1 |

\(2^1\) | 01 | 2 | 2 |

\(2^2\) | 100 | 4 | 4 |

\(2^3\) | 1000 | 8 | 8 |

\(2^4\) | 1 0000 | 16 | 10 |

\(2^5\) | 10 0000 | 32 | 20 |

\(2^6\) | 100 0000 | 64 | 40 |

\(2^7\) | 1000 0000 | 128 | 80 |

\(2^8\) | 1 0000 0000 | 256 | 1 00 |

\(2^9\) | 10 0000 0000 | 512 | 2 00 |

\(2^{10}\) | … | (1K) \(1,024\) | 4 00 |

\(2^{11}\) | \(2,048\) | 8 00 | |

\(2^{12}\) | \(4,096\) | 10 00 | |

\(2^{13}\) | \(8,192\) | 20 00 | |

\(2^{14}\) | \(16,384\) | 40 00 | |

\(2^{15}\) | \(32,768\) | 80 00 | |

\(2^{16}\) | \(65,536\) | 1 00 00 | |

\(2^{17}\) | \(131,072\) | 2 00 00 | |

\(2^{18}\) | \(262,144\) | 4 00 00 | |

\(2^{19}\) | \(524,288\) | 8 00 00 | |

\(2^{20}\) | (1 MB) \(1,048,576\) | 10 00 00 | |

\(2^{21}\) | \(2,097,152\) | 20 00 00 | |

\(2^{22}\) | \(4,194,304\) | 40 00 00 | |

\(2^{23}\) | \(8,388,608\) | 80 00 00 | |

\(2^{24}\) | \(16,777,216\) | 1 00 00 00 | |

\(2^{25}\) | \(33,554,432\) | 2 00 00 00 | |

\(2^{26}\) | \(67,108,864\) | 4 00 00 00 | |

\(2^{27}\) | \(134,217,728\) | 8 00 00 00 | |

\(2^{28}\) | \(268,435,456\) | 10 00 00 00 | |

\(2^{29}\) | \(536,870,912\) | 20 00 00 00 | |

\(2^{30}\) | (1 GB) \(1,073,741,824\) | 40 00 00 00 | |

\(2^{31}\) | \(2,147,483,648\) | 80 00 00 00 | |

\(2^{32}\) | \(4,294,967,296\) | 1 00 00 00 00 | |

\(2^{33}\) | \(8,589,934,592\) | 2 00 00 00 00 | |

\(2^{34}\) | \(17,179,869,184\) | 4 00 00 00 00 | |

\(2^{35}\) | \(34,359,738,368\) | 8 00 00 00 00 | |

\(2^{36}\) | \(68,719,476,736\) | 10 00 00 00 00 | |

\(2^{37}\) | \(137,438,953,472\) | 20 00 00 00 00 | |

\(2^{38}\) | \(274,877,906,944\) | 40 00 00 00 00 | |

\(2^{39}\) | \(549,755,813,888\) | 80 00 00 00 00 | |

\(2^{40}\) | (1 tera) \(1,099,511,627,776\) | 1 00 00 00 00 00 | |

\(2^{41}\) | \(2,199,023,255,552\) | 2 00 00 00 00 00 | |

\(2^{42}\) | \(4,398,046,511,104\) | 4 00 00 00 00 00 | |

\(2^{43}\) | \(8,796,093,022,208\) | 8 00 00 00 00 00 | |

\(2^{44}\) | \(17,592,186,044,416\) | 10 00 00 00 00 00 | |

\(2^{45}\) | \(35,184,372,088,832\) | 20 00 00 00 00 00 | |

\(2^{46}\) | \(70,368,744,177,664\) | 40 00 00 00 00 00 | |

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

\(2^{47}\) | 100000… | \(140,737,488,355,328\) | 80 00 00 00 00 00 |

\(2^{48}\) | \(281,474,976,710,656\) | 1 00 00 00 00 00 00 | |

\(2^{49}\) | \(562,949,953,421,312\) | 2 00 00 00 00 00 00 | |

\(2^{50}\) | \(1,125,899,906,842,624\) | 4 00 00 00 00 00 00 | |

\(2^{51}\) | \(2,251,799,813,685,248\) | 8 00 00 00 00 00 00 | |

\(2^{52}\) | \(4,503,599,627,370,496\) | 10 00 00 00 00 00 00 | |

\(2^{53}\) | \(9,007,199,254,740,992\) | 20 00 00 00 00 00 00 | |

\(2^{54}\) | \(18,014,398,509,481,984\) | 40 00 00 00 00 00 00 | |

\(2^{55}\) | \(36,028,797,018,963,968\) | 80 00 00 00 00 00 00 | |

\(2^{56}\) | \(72,057,594,037,927,936\) | 1 00 00 00 00 00 00 00 | |

\(2^{57}\) | \(144,115,188,075,855,872\) | 2 00 00 00 00 00 00 00 | |

\(2^{58}\) | \(288,230,376,151,711,744\) | 4 00 00 00 00 00 00 00 | |

\(2^{59}\) | \(576,460,752,303,423,488\) | 8 00 00 00 00 00 00 00 | |

\(2^{60}\) | \(1,152,921,504,606,846,976\) | 10 00 00 00 00 00 00 00 | |

\(2^{61}\) | \(2,305,843,009,213,693,952\) | 20 00 00 00 00 00 00 00 | |

\(2^{62}\) | \(4,611,686,018,427,387,904\) | 40 00 00 00 00 00 00 00 | |

\(2^{63}\) | \(9,223,372,036,854,775,808\) | 80 00 00 00 00 00 00 00 | |

\(2^{64}\) | \(18,446,744,073,709,551,616\) | 1 00 00 00 00 00 00 00 00 | |

Java uses IEEE 754.

A number such as \(0.125 \) is expressed as \(1.25 \cdot 10^{-1}\) or \(1 \cdot 2^{-3}\).

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.

\begin{eqnarray} s \cdot m \cdot 2^{E-N+1} \nonumber \\ s \mbox{ is the sign, and can be} -1 \mbox{or} +1 \nonumber \\ 1 \le m \le 2^{24}-1 = 16,777,215 \nonumber \\ -126 \le E \le +127 \nonumber \nonumber \\ N=24 \nonumber \end{eqnarray}

So, from the above, a ﬂoat \(f\) in IEEE 754 is in the range

\begin{eqnarray} -1 \cdot 16777215 \cdot 2^{ -126 - 24 +1} \le f \le +1 \cdot 16777215 \cdot 2^{127 - 24 +1} \nonumber \\ -16777215 \cdot 2^{ -149} \le f \le +16777215 \cdot 2^{104} \nonumber \\ -2.35 \cdot 10^{-38} \le f \le 3.4 \cdot 10^{38} \nonumber \end{eqnarray}

In Java a double is expressed as

\begin{eqnarray} s \cdot m \cdot 2^{E-N+1} \nonumber \\ s \mbox{ is the sign, and can be} -1 \mbox{or} +1 \nonumber \\ 1 \le m \le 2^{53}-1 = 9,007,199,254,740,991 \nonumber \\ -1022 \le E \le +1023 \nonumber \nonumber \\ N=24 \nonumber \end{eqnarray}

So, from the above, a double \(f\) in IEEE 754 is in the range

\begin{eqnarray} -1 \cdot 9007199254740991 \cdot 2^{ -1022 - 24 +1} \le f \le +1 \cdot 9007199254740991 \cdot 2^{1023 - 24 +1} \nonumber \\ -9007199254740991 \cdot 2^{ -1045} \le f \le +9007199254740991 \cdot 2^{1000} \nonumber \\ -2.2 \cdot 10^{-308} \le f \le 1.8 \cdot 10^{308} \nonumber \end{eqnarray}

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 \(2^{8}\). 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:

\(1 + 0(1/2) + 0(1/4) + 1(1/8) + 0(1/16) + 1(1/32) + 1(1/64) + 0(1/128) + 0(1/256) + \ldots all zeros\)

which is \( 1+(1/8)+(1/32)+(1/64) = 1+(11/64) = 75/64\)

Hence the ﬁnal number is \(-(75/64) \cdot 2^{8} = -(75/64) \cdot 256 = -300\).

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 \(m \cdot 2^e\), assume \(e=0\), then the accuracy of a ﬂoat goes like this: \(1, 1+(1/2), 1+(1/2)+(1/4), 1+(1/2)+(1/4)+(1/8), 1+(1/2)+(1/4)+(1/8)+(1/16), \ldots \) or \(1, 1.5, 1.75, 1.87, \ldots \),

So, a number such as \(1.4\) can’t be exactly expressed in ﬂoating point ! because the \(.4\) 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 \((2^{128} - 2^{104})\), This is 40 digits number, which is represented by \(01111111011111111111111111111111\)

The Java programing language speciﬁcations.

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