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1 Java primitive types sizes

2 Maximum value in signed and unsigned integers

Signed integer table

Unsigned integer table

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.

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

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.

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

In Java a double is expressed as

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

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