How to solve multiple decimal places in javascript decimal operations

I have studied how to add, subtract, multiply, and divide JavaScript decimal places. I have also done some tests and gained a lot of knowledge, you can share an interesting test with us as follows:

0.1 + 0.2 = 0.3 // false
Suddenly depressed, okay! 0.1 + 0.2 changed to: 0.30000000000000004
Another 2.4/0.8 => 2.9999999999999996 cannot be converted into an integer (2.4*100)/(0.8*100)
10.22 now I want to subtract the result value of 0.11 and then there will be a lot of decimal places 10.110000000000001. Then I used the toFixed method to filter decimal places, but I don't know which efficiency is high to convert it to an integer! Let's try again!

Var date1 = new Date (); for (var I = 0; I <10000; I ++) {var result1 = (10.22-0.11 ). toFixed (2);} alert (new Date ()-date1); // low efficiency var date2 = new Date (); for (var j = 0; j <10000; j ++) {var result2 = (10.22*1000-0.11*1000)/1000;} alert (new Date ()-date2 ); // High Efficiency alert (0.1 + 0.2 = 0.3); // Since falsealert (0.1 + 0.2) is returned; // Since 0.30000000000000004 alert (parseFloat (0.1) is returned) + parseFloat (0.2); // returns 0.30000000000000004

I checked some information. One is the Bug of JavaScript floating-point number calculation, and the other is related to the computer's final conversion to binary calculation. But why not all decimals will have this problem? I am not sure yet, I have time to study it in depth.

Solution:
To solve this problem, the first method is to use the toFixed (n) method of JavaScript to directly obtain N decimal places. However, I think this method has some problems in data precision. If the data accuracy requirement is not high, you can use it.

alert((0.1 + 0.2).toFixed(1));

The second method is to write the calculation method by yourself. The following is a user-defined addition function. Using this method for addition will avoid the above problem.

// UDF addNum (num1, num2) {var sq1, sq2, m; try {sq1 = num1.toString (). split (". ") [1]. length;} catch (e) {sq1 = 0;} try {sq2 = num2.toString (). split (". ") [1]. length;} catch (e) {sq2 = 0;} m = Math. pow (10, Math. max (sq1, sq2); return (num1 * m + num2 * m)/m;} alert (addNum (0.1, 0.2 ));

Of course, you can also write it as follows: alert (num * 3 + 10*3)/3). N decimal places are not displayed.
Alert (num * 3 + 10*3)/3); and alert (num + 10); There is a difference between the two programming languages: Converting computers to binary operations at the underlying layer, perhaps this is the reason for the above problem. We will discuss it further.