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Java floating point numbers review

Nasser M. Abbasi

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

Contents

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

1 Java primitive types sizes



type size in bytes


byte 1
short 2
int 4
long 8
float 4 (IEEE 754)
double 8 (IEEE 754)


2 Maximum value in signed and unsigned integers

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 \(\ldots 00 00 00 00 00 00 00 00\)




3 Some bits table

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.

4 Power of 2 table





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




5 Float and Double in Java

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}

5.1 How to read a floating point?

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 \(01111111011111111111111111111111\)

6 References

The Java programing language specifications.

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