Java float data storage, loss of precision problems

The float type in Java is 4 bytes 32 bits, and in-memory storage follows the IEEE-754 format standard:

A floating-point number consists of 2 parts: base m and index E
Base m part: uses a binary number to represent the actual value of this floating point.
Index e part: A binary number that occupies 8bit (1 bytes), which can represent a range of values of 0-255.
However, the index can be negative, so the IEEE stipulates that the sub-square must be subtracted from 127 to be the true exponent.
Therefore, the float type index can be from 126 to 128.
The base portion is actually a value of 24bit (3 bytes), but the highest level is always 1, so the top bit is omitted from storage, which accounts for 23bit in storage.

So 32 bits of a floating-point number are made up of three parts:

1. The first bit is the sign bit, indicating whether the floating point number is a positive or negative, 0 is positive, and 1 is negative;

2. The second to the Nineth place a total of 8 to denote the exponent E, turns into the binary need +127 (conversely minus 127);

3. The last 23 bits are the base, but the storage automatically omits the highest 1, so the storage is 23 bits actually 24 bits. (Decimal position is 2^ (23) =8388608 7 digits calculated as 6-digit precision)

For a chestnut: 3.2f

Convert to binary storage: integral part 32 binary 11

Fractional part 0.2 is converted to binary (by 2 to the integer portion until the fractional part is divided into 0)----> (decimal by 2) 0.4 0.8 1.6 1.2 0.4 0.8 1.6 1.2 0.4 ... (Infinite Loop)

The result is: 11.00 1100 11001100 11001100 11001100 (because of the limited number of bits, and the fractional part is infinitely bad, this has lost precision)

Move right until you have only one left: 1.100 1100 11001100 11001100 11001100 (one bit to the right, the exponential portion is 1+127=128 1000 0000)

Finally get binary storage: 0 100 0000 0100 1100 11001100 11001100 (loss of precision)

Inverse conversion to floating point decimal: Sign bit 0: positive

Index 100 0000 0 is 128 minus 127 index is 1

Number of parts: the highest bit 1 is omitted by default plus the decimal point is 1.100 1100 11001100 11001100

Positive index right Shift one 11.0011 00110011 0011001100

Positive number part---->3 Base part 2^ (-3) +2^ (-4) +2^ (-7) +2^ (-8) +2^ (-11) +2^ (-12) +2^ (-15) +2^ (-16) +2^ (-19) +2^ (-20)

Java float data storage, loss of precision problems