The problem of JS arithmetic decimal point

Problem like this: 37.5*5.5=206.08 (JS figure out is such a result, I rounded to take two decimal places) I first suspect is rounding problem, directly with JS forget a result is: 206.08499999999998 how can this, Two numbers with only one decimal number multiply, how can the decimal point come out more so.

I Google a bit, found that this is a JavaScript floating point operation of a bug. For example: 7*0.8 JavaScript is calculated to be: 5.6000000000000005 on the internet to find some solutions, is to re-write some of the functions of floating-point operations. Here are some excerpts from these methods for the reference of friends who encounter the same problem:

Program code//Division function 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

function Accdiv (ARG1,ARG2) {

var t1=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 (10,T2-T1); } }

Adding a Div method to the number type makes it easier to call.

Number.prototype.div = function (ARG) {

Return Accdiv (this, ARG);

}

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

function Accmul (ARG1,ARG2) {

var m=0,s1=arg1.tostring (), s2=arg2.tostring ();

Try{m+=s1.split (".") [1].length}catch (e) {}

Try{m+=s2.split (".") [1].length}catch (e) {}

Return number (S1.replace (".", "")) *number (S2.replace (".", ""))/math.pow (10,m)}

Adding a Mul method to the number type makes it more convenient to call.

Number.prototype.mul = function (ARG) {return Accmul (ARG, this);}

The addition function is used to get the exact addition//description: 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 Precision results

function Accadd (ARG1,ARG2) {

var r1,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

}

Adding an Add method to the number type makes it more convenient to call.

Number.prototype.add = function (ARG) {

Return Accadd (Arg,this);

}

Include these functions where you want them, and then call it to calculate. For example you want to calculate: 7*0.8, then change to (7). Mul (8) The other operations are similar, so that more accurate results can be obtained. --------------------------------------------------------------------------------------------------------------- ---------------------------------------------above is reproduced on the internet a JS cattle blog, but the above only mention the solution of addition, multiplication and division. At this time, many people will think, with addition, subtraction is not easy? I just almost let this idea suffer. The rest is not much to say, the code of the subtraction:

function Subtr (ARG1,ARG2) {

var r1,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 length

N= (R1>=R2) r1:r2;

Return ((arg1*m-arg2*m)/m). ToFixed (n); }

The problem of JS arithmetic decimal point