/****************************************解决JS浮点数(小数)计算加减乘除的BUG ???Start****************************************//**** 加法函数，用来得到精确的加法结果** 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。** 调用：accAdd(arg1,arg2)** 返回值：arg1加上arg2的精确结果**/function accAdd(arg1, arg2) { ???if (isNaN(arg1)) { ???????arg1 = 0; ???} ???if (isNaN(arg2)) { ???????arg2 = 0; ???} ???arg1 = Number(arg1); ???arg2 = Number(arg2); ???var r1, r2, m, c; ???try { ???????r1 = arg1.toString().split(".")[1].length; ???} ???catch (e) { ???????r1 = 0; ???} ???try { ???????r2 = arg2.toString().split(".")[1].length; ???} ???catch (e) { ???????r2 = 0; ???} ???c = Math.abs(r1 - r2); ???m = Math.pow(10, Math.max(r1, r2)); ???if (c > 0) { ???????var cm = Math.pow(10, c); ???????if (r1 > r2) { ???????????arg1 = Number(arg1.toString().replace(".", "")); ???????????arg2 = Number(arg2.toString().replace(".", "")) * cm; ???????} else { ???????????arg1 = Number(arg1.toString().replace(".", "")) * cm; ???????????arg2 = Number(arg2.toString().replace(".", "")); ???????} ???} else { ???????arg1 = Number(arg1.toString().replace(".", "")); ???????arg2 = Number(arg2.toString().replace(".", "")); ???} ???return (arg1 + arg2) / m;}//给Number类型增加一个add方法，调用起来更加方便。Number.prototype.add = function(arg) { ???return accAdd(this, arg);};/**** 减法函数，用来得到精确的减法结果** 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。** 调用：accSub(arg1,arg2)** 返回值：arg1加上arg2的精确结果**/function accSub(arg1, arg2) { ???if (isNaN(arg1)) { ???????arg1 = 0; ???} ???if (isNaN(arg2)) { ???????arg2 = 0; ???} ???arg1 = Number(arg1); ???arg2 = Number(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 //动态控制精度长度 ???n = (r1 >= r2) ? r1 : r2; ???return ((arg1 * m - arg2 * m) / m).toFixed(n);}// 给Number类型增加一个mul方法，调用起来更加方便。Number.prototype.sub = function(arg) { ???return accSub(this, arg);};/**** 乘法函数，用来得到精确的乘法结果** 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。** 调用：accMul(arg1,arg2)** 返回值：arg1乘以 arg2的精确结果**/function accMul(arg1, arg2) { ???if (isNaN(arg1)) { ???????arg1 = 0; ???} ???if (isNaN(arg2)) { ???????arg2 = 0; ???} ???arg1 = Number(arg1); ???arg2 = Number(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);}// 给Number类型增加一个mul方法，调用起来更加方便。Number.prototype.mul = function(arg) { ???return accMul(this, arg);};/** ** 除法函数，用来得到精确的除法结果** 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。** 调用：accDiv(arg1,arg2)** 返回值：arg1除以arg2的精确结果**/function accDiv(arg1, arg2) { ???if (isNaN(arg1)) { ???????arg1 = 0; ???} ???if (isNaN(arg2)) { ???????arg2 = 0; ???} ???arg1 = Number(arg1); ???arg2 = Number(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); ???}}//给Number类型增加一个div方法，调用起来更加方便。Number.prototype.div = function(arg) { ???return accDiv(this, arg);};/****************************************解决JS浮点数(小数)计算加减乘除的BUG ?End****************************************/解决JS浮点数(小数)计算加减乘除的BUG
原文地址：https://www.cnblogs.com/whboxl/p/8342570.html