Solve the problem of loss of computational precision in JavaScript

In the project before the teacher gave us a package of a JS file, to solve the loss of precision in the calculation of some functions, direct reference to the JS file can be used.

eg

var NumA = 0.1;
var NumB = 0.2;
Alert (NumA + NumB);

Results occurred: 0.1 + 0.2 = 0.30000000000000004

Why this problem occurs: the computer reads the binary, not the decimal, is the program in the conversion time lost precision.

To resolve the problem code:

    //The division function, which is used to get accurate division results.    //Description: The result of the division of JavaScript will be error, which will be obvious when dividing two floating-point numbers. This function returns a more accurate division result.     //Call: Accdiv (ARG1,ARG2)    //return value: Arg1 divided by the exact result of Arg2    functionAccdiv (arg1, arg2) {varT1 = 0, t2 = 0, R1, R2; Try{T1 = arg1.tostring (). Split (".") [1].length}Catch(e) {}Try{t2 = arg2.tostring (). Split (".") [1].length}Catch(e) {} with(Math) {R1= Number (arg1.tostring (). Replace (".", "")); R2= Number (arg2.tostring (). Replace (".", "")); return(R1/R2) * POW (T2-t1); }    }    //multiplication function to get the exact multiplication result    //Description: JavaScript multiplication results are error-evident when multiplying two floating-point numbers. This function returns a more accurate multiplication result.     //Call: Accmul (ARG1,ARG2)    //return value: arg1 times the exact result of Arg2    functionAccmul (arg1, arg2) {varm = 0, S1 = arg1.tostring (), S2 =arg2.tostring (); Try{m + = S1.split (".") [1].length}Catch(e) {}Try{m + = S2.split (".") [1].length}Catch(e) {}returnNumber (S1.replace (".", "")) * Number (S2.replace (".", ""))/Math.pow (10, M); }    //An addition function that is used to obtain accurate addition results    //Note: The addition of JavaScript will have an error, and it will be more obvious when the two floating-point numbers are added. This function returns a more accurate addition result.     //Call: Accadd (ARG1,ARG2)    //return value: Arg1 plus arg2 's exact result    functionAccadd (arg1, arg2) {varR1, R2, M; Try{r1 = arg1.tostring (). Split (".") [1].length; }Catch(e) {r1 = 0; } Try{r2 = arg2.tostring (). Split (".") [1].length; }Catch(e) {r2 = 0; } M= Math.pow (10, Math.max (R1, r2)); return(ARG1 * m + arg2 * m)/m; }    //Subtraction Function    functionaccsub (arg1, arg2) {varR1, R2, M, N; Try{R1= Arg1.tostring (). Split (".") [1].length; } Catch(e) {R1= 0; }        Try{R2= Arg2.tostring (). Split (".") [1].length; } Catch(e) {R2= 0; } m= Math.pow (10, Math.max (R1, r2)); //Last modify by Deeka        //Dynamic Control accuracy lengthn = (r1 >= r2)?R1:R2; return((ARG2 * M-ARG1 * m)/m). toFixed (n); }            //Adding an Add method to the number type makes it more convenient to call. Number.prototype.add =function(ARG) {returnAccadd (ARG, This);    }; //Adding a sub method to the number class makes it easier to call.Number.prototype.sub =function(ARG) {returnAccsub (ARG, This);    }; //Add a Mul method to the number typeNumber.prototype.mul =function(ARG) {returnAccmul (ARG, This);    }; //extend a Div method to the number typeNumber.prototype.div =function(ARG) {returnAccdiv ( This, ARG); };

Solve the problem of loss of computational precision in JavaScript