JS - accuracy issues

I'm learning today js In the process , A precision problem was found 2^53 -1

To find out

1. According to international standards IEEE 754,javascript Floating point 64 Binary bits , From the far left , It's made up of
   The first 1 position ： Sign bit ,0 It means a positive number ,1 A negative number (1 position )
   The first 2 Position to the first 12 position ： Storage index section (11 position )
   The first 13 Position to the first 64 position ： Save decimal part （ It's a significant number 52 position , remove Sign bit and exponent bit ）

The sign bit determines the plus or minus of a number , The exponential part determines the size of the number , The decimal part determines the numerical precision

IEEE 754 Regulations , The first significant number by default is always 1, Don't save it in 64 Bit floating point number .
 in other words , Significant numbers are always 1.xx...xx In the form of , among xx..xx Part of the is saved in 64 Bit floating point number , The longest possible is 52 position 

therefore ,javascript The most significant number provided is 53 Binary bits
The maximum accuracy is 53 Binary bits , Means that the absolute value is less than 2 Of 53 The whole number of the power , namely ±(2^53-1), Can accurately represent

Look at a few demo

0.1+0.2!=0.3
0.1 Expressed as binary
0.0001100110011001100110011001100110011001100110011001101
0.2 Expressed as binary
0.001100110011001100110011001100110011001100110011001101
Add up in the computer , just
0.0100110011001100110011001100110011001100110011001100111
To convert to decimal is 0.30000000000000004

Explain why other situations are similar to 0.9+0.1==1 Well ？
This is because although the added binary number is not exactly equal to 1, But the extra value 1.0000000000000000000000000000000000000000000000000000001
Is too small , It was abandoned

resolvent

If the accuracy requirement is not high , direct (0.1+0.2).toFixed(2) == 0.3
utilize ES6 Number new property EPSILON Math.abs(0.1+0.2-0.3)<Number.EPSILON
Utilization Library mathjs,math.format(math.eval(‘0.1 + 0.2’))