解决JS浮点数(小数)计算加减乘除的BUG

在JavaScript中输出下面这些数值(注意不能作为字符串输出):0.1000000000000000000000000001(28位小数)、0.100000000000000000000000001(27位小数)、0.1000000000000000000000000456(28位小数)、0.09999999999999999999999(23位小数),显示出来的结果都是数值0.1。又如,如果输出1/3的有理数表达式,结果是0.3333333333333333。 document.write(.1 + .2) // 0.3000000000000004

document.write(.3 + .6) // 0.8999999999999999

Some statistics related to this famous double precision question. I used this code.

When adding all values (a+b) using a step of 0.1 (from 0.1 to 100) we have ~15% chance of precision error. Here are some examples (for full .txt results here):

0.1 + 0.2 = 0.30000000000000004
0.1 + 0.7 = 0.7999999999999999
...
1.7 + 1.9 = 3.5999999999999996
1.7 + 2.2 = 3.9000000000000004
...
3.2 + 3.6 = 6.800000000000001
3.2 + 4.4 = 7.6000000000000005

When subtracting all values (a-b where a>b) using a step of 0.1 (from 100 to 0.1) we have ~34% chance of precision error. Here are some examples (for full .txt results here):

0.6 - 0.2 = 0.39999999999999997
0.5 - 0.4 = 0.09999999999999998
...
2.1 - 0.2 = 1.9000000000000001
2.0 - 1.9 = 0.10000000000000009
...
100 - 99.9 = 0.09999999999999432
100 - 99.8 = 0.20000000000000284

*I was surprised with these 15% and 34%.. they are huge, so always use BigDecimal when precision is of big importance. With 2 decimal digits (step 0.01) the situation worsens a bit more (18% and 36%).

 
解决JS浮点数(小数)计算加减乘除的BUG
 1 /**
2 ** 加法函数,用来得到精确的加法结果
3 ** 说明:javascript的加法结果会有误差,在两个浮点数相加的时候会比较明显。这个函数返回较为精确的加法结果。
4 ** 调用:accAdd(arg1,arg2)
5 ** 返回值:arg1加上arg2的精确结果
6 **/
7 function accAdd(arg1, arg2) {
8 var r1, r2, m, c;
9 try {
10 r1 = arg1.toString().split(".")[1].length;
11 }
12 catch (e) {
13 r1 = 0;
14 }
15 try {
16 r2 = arg2.toString().split(".")[1].length;
17 }
18 catch (e) {
19 r2 = 0;
20 }
21 c = Math.abs(r1 - r2);
22 m = Math.pow(10, Math.max(r1, r2));
23 if (c > 0) {
24 var cm = Math.pow(10, c);
25 if (r1 > r2) {
26 arg1 = Number(arg1.toString().replace(".", ""));
27 arg2 = Number(arg2.toString().replace(".", "")) * cm;
28 } else {
29 arg1 = Number(arg1.toString().replace(".", "")) * cm;
30 arg2 = Number(arg2.toString().replace(".", ""));
31 }
32 } else {
33 arg1 = Number(arg1.toString().replace(".", ""));
34 arg2 = Number(arg2.toString().replace(".", ""));
35 }
36 return (arg1 + arg2) / m;
37 }
38
39 //给Number类型增加一个add方法,调用起来更加方便。
40 Number.prototype.add = function (arg) {
41 return accAdd(arg, this);
42 };
解决JS浮点数(小数)计算加减乘除的BUG
解决JS浮点数(小数)计算加减乘除的BUG
 1 /**
2 ** 减法函数,用来得到精确的减法结果
3 ** 说明:javascript的减法结果会有误差,在两个浮点数相减的时候会比较明显。这个函数返回较为精确的减法结果。
4 ** 调用:accSub(arg1,arg2)
5 ** 返回值:arg1加上arg2的精确结果
6 **/
7 function accSub(arg1, arg2) {
8 var r1, r2, m, n;
9 try {
10 r1 = arg1.toString().split(".")[1].length;
11 }
12 catch (e) {
13 r1 = 0;
14 }
15 try {
16 r2 = arg2.toString().split(".")[1].length;
17 }
18 catch (e) {
19 r2 = 0;
20 }
21 m = Math.pow(10, Math.max(r1, r2)); //last modify by deeka //动态控制精度长度
22 n = (r1 >= r2) ? r1 : r2;
23 return ((arg1 * m - arg2 * m) / m).toFixed(n);
24 }
25
26 // 给Number类型增加一个mul方法,调用起来更加方便。
27 Number.prototype.sub = function (arg) {
28 return accMul(arg, this);
29 };
解决JS浮点数(小数)计算加减乘除的BUG
解决JS浮点数(小数)计算加减乘除的BUG
 1 /**
2 ** 乘法函数,用来得到精确的乘法结果
3 ** 说明:javascript的乘法结果会有误差,在两个浮点数相乘的时候会比较明显。这个函数返回较为精确的乘法结果。
4 ** 调用:accMul(arg1,arg2)
5 ** 返回值:arg1乘以 arg2的精确结果
6 **/
7 function accMul(arg1, arg2) {
8 var m = 0, s1 = arg1.toString(), s2 = arg2.toString();
9 try {
10 m += s1.split(".")[1].length;
11 }
12 catch (e) {
13 }
14 try {
15 m += s2.split(".")[1].length;
16 }
17 catch (e) {
18 }
19 return Number(s1.replace(".", "")) * Number(s2.replace(".", "")) / Math.pow(10, m);
20 }
21
22 // 给Number类型增加一个mul方法,调用起来更加方便。
23 Number.prototype.mul = function (arg) {
24 return accMul(arg, this);
25 };
解决JS浮点数(小数)计算加减乘除的BUG
解决JS浮点数(小数)计算加减乘除的BUG
 1 /**
2 ** 除法函数,用来得到精确的除法结果
3 ** 说明:javascript的除法结果会有误差,在两个浮点数相除的时候会比较明显。这个函数返回较为精确的除法结果。
4 ** 调用:accDiv(arg1,arg2)
5 ** 返回值:arg1除以arg2的精确结果
6 **/
7 function accDiv(arg1, arg2) {
8 var t1 = 0, t2 = 0, r1, r2;
9 try {
10 t1 = arg1.toString().split(".")[1].length;
11 }
12 catch (e) {
13 }
14 try {
15 t2 = arg2.toString().split(".")[1].length;
16 }
17 catch (e) {
18 }
19 with (Math) {
20 r1 = Number(arg1.toString().replace(".", ""));
21 r2 = Number(arg2.toString().replace(".", ""));
22 return (r1 / r2) * pow(10, t2 - t1);
23 }
24 }
25
26 //给Number类型增加一个div方法,调用起来更加方便。
27 Number.prototype.div = function (arg) {
28 return accDiv(this, arg);
29 };
解决JS浮点数(小数)计算加减乘除的BUG
 
高级浏览器解决方法:
 
function strip(number) {
return (parseFloat(number.toPrecision(12)));
}

Using 'toPrecision(12)' leaves trailing zeros which 'parseFloat()' removes. Assume it is accurate to plus/minus one on the least significant digit.

定义和用法

toPrecision() 方法可在对象的值超出指定位数时将其转换为指数计数法。

语法

NumberObject.toPrecision(num)
参数 描述
num 必需。规定必须被转换为指数计数法的最小位数。该参数是 1 ~ 21 之间(且包括 1 和 21)的值。有效实现允许有选择地支持更大或更小的 num。如果省略了该参数,则调用方法 toString(),而不是把数字转换成十进制的值。

返回值

返回 NumberObject 的字符串表示,包含 num 个有效数字。如果 num 足够大,能够包括 NumberObject 整数部分的所有数字,那么返回的字符串将采用定点计数法。否则,采用指数计数法,即小数点前有一位数字,小数点后有 num-1 位数字。必要时,该数字会被舍入或用 0 补足。

抛出

当 num 太小或太大时抛出异常 RangeError。1 ~ 21 之间的值不会引发该异常。有些实现支持更大范围或更小范围内的值。

当调用该方法的对象不是 Number 时抛出 TypeError 异常。

实例

在本例中,我们将把一个数字转换为指数计数法:

Show 10,000 as an exponential notation:
<script type="text/javascript">
var num = new Number(10000);
document.write (num.toPrecision(4))
</script>

输出:

Show 10,000 as an exponential notation:
1.000e+4 同时还可以使用一个https://github.com/MikeMcl/bignumber.js 插件来完成JS精度运算的BUG

Faster, smaller, and perhaps easier to use than JavaScript versions of Java's BigDecimal
8 KB minified and gzipped
Simple API but full-featured
Works with numbers with or without fraction digits in bases from 2 to 64 inclusive
Replicates the toExponential, toFixed, toPrecision and toString methods of JavaScript's Number type
Includes a toFraction and a correctly-rounded squareRoot method
Supports cryptographically-secure pseudo-random number generation
No dependencies
Wide platform compatibility: uses JavaScript 1.5 (ECMAScript 3) features only
Comprehensive documentation and test set
API

If a smaller and simpler library is required see big.js. It's less than half the size but only works with decimal numbers and only has half the methods. It also does not allow NaN or Infinity, or have the configuration options of this library.

Use

In all examples below, var, semicolons and toString calls are not shown. If a commented-out value is in quotes it means toString has been called on the preceding expression.

The library exports a single function: BigNumber, the constructor of BigNumber instances.

It accepts a value of type number (up to 15 significant digits only), string or BigNumber object,

x = new BigNumber(123.4567)
y = BigNumber('123456.7e-3')
z = new BigNumber(x)
x.equals(y) && y.equals(z) && x.equals(z) // true
and a base from 2 to 64 inclusive can be specified.

x = new BigNumber(1011, 2) // "11"
y = new BigNumber('zz.9', 36) // "1295.25"
z = x.plus(y) // "1306.25"
A BigNumber is immutable in the sense that it is not changed by its methods.

0.3 - 0.1 // 0.19999999999999998
x = new BigNumber(0.3)
x.minus(0.1) // "0.2"
x // "0.3"
The methods that return a BigNumber can be chained.

x.dividedBy(y).plus(z).times(9).floor()
x.times('1.23456780123456789e+9').plus(9876.5432321).dividedBy('4444562598.111772').ceil()
Many method names have a shorter alias.

x.squareRoot().dividedBy(y).toPower(3).equals(x.sqrt().div(y).pow(3)) // true
x.cmp(y.mod(z).neg()) == 1 && x.comparedTo(y.modulo(z).negated()) == 1 // true
Like JavaScript's number type, there are toExponential, toFixed and toPrecision methods

x = new BigNumber(255.5)
x.toExponential(5) // "2.55500e+2"
x.toFixed(5) // "255.50000"
x.toPrecision(5) // "255.50"
x.toNumber() // 255.5
and a base can be specified for toString.

x.toString(16) // "ff.8"
There is also a toFormat method which may be useful for internationalisation

y = new BigNumber('1234567.898765')
y.toFormat(2) // "1,234,567.90"
The maximum number of decimal places of the result of an operation involving division (i.e. a division, square root, base conversion or negative power operation) is set using the config method of the BigNumber constructor.

The other arithmetic operations always give the exact result.

BigNumber.config({ DECIMAL_PLACES: 10, ROUNDING_MODE: 4 })
// Alternatively, BigNumber.config( 10, 4 );

x = new BigNumber(2);
y = new BigNumber(3);
z = x.div(y) // "0.6666666667"
z.sqrt() // "0.8164965809"
z.pow(-3) // "3.3749999995"
z.toString(2) // "0.1010101011"
z.times(z) // "0.44444444448888888889"
z.times(z).round(10) // "0.4444444445"
There is a toFraction method with an optional maximum denominator argument

y = new BigNumber(355)
pi = y.dividedBy(113) // "3.1415929204"
pi.toFraction() // [ "7853982301", "2500000000" ]
pi.toFraction(1000) // [ "355", "113" ]
and isNaN and isFinite methods, as NaN and Infinity are valid BigNumber values.

x = new BigNumber(NaN) // "NaN"
y = new BigNumber(Infinity) // "Infinity"
x.isNaN() && !y.isNaN() && !x.isFinite() && !y.isFinite() // true
The value of a BigNumber is stored in a decimal floating point format in terms of a coefficient, exponent and sign.

x = new BigNumber(-123.456);
x.c // [ 123, 45600000000000 ] coefficient (i.e. significand)
x.e // 2 exponent
x.s // -1 sign
Multiple BigNumber constructors can be created, each with their own independent configuration which applies to all BigNumber's created from it.

// Set DECIMAL_PLACES for the original BigNumber constructor
BigNumber.config({ DECIMAL_PLACES: 10 })

// Create another BigNumber constructor, optionally passing in a configuration object
BN = BigNumber.another({ DECIMAL_PLACES: 5 })

x = new BigNumber(1)
y = new BN(1)

x.div(3) // '0.3333333333'
y.div(3) // '0.33333'
For futher information see the API reference in the doc directory.

 
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